Kinematics - Topical Past Papers
1.
[Source: 0606/22/F/M/15 - Question No. 11]
- A particle $P$ moves in a straight line. Starting from rest, $P$ moves with constant acceleration for 30 seconds after which it moves with constant velocity, $\displaystyle k \text{ ms}^{-1}$, for 90 seconds. $P$ then moves with constant deceleration until it comes to rest; the magnitude of the deceleration is twice the magnitude of the initial acceleration.
- Use the information to complete the velocity-time graph.
- Given that the particle travels $450$ metres while it is accelerating, find the value of $\displaystyle k$ and the acceleration of the particle.
- Use the information to complete the velocity-time graph.
- A body moves in a straight line such that, $\displaystyle t$ seconds after passing a fixed point, its acceleration, $\displaystyle a \text{ ms}^{-2}$, is given by $\displaystyle a=3t^{2}+6$. When $\displaystyle t=0$, the velocity of the body is $\displaystyle 5 \text{ ms}^{-1}$. Find the velocity when $\displaystyle t=3$.
(a)(i)
From $\displaystyle t = 0$ to $\displaystyle t = 30$, velocity increases linearly to $\displaystyle k$.
From $\displaystyle t = 30$ to $\displaystyle t = 120$ ($30 + 90$), velocity remains constant at $\displaystyle k$.
Let initial acceleration be $\displaystyle a_1 = \displaystyle \frac{k}{30}$.
Deceleration magnitude is $\displaystyle 2a_1 = \displaystyle \frac{2k}{30} = \frac{k}{15}$.
Time taken to come to rest from velocity $\displaystyle k$: $\displaystyle \Delta t = \frac{k}{k/15} = 15$ seconds.
The graph goes down to $\displaystyle v=0$ at $\displaystyle t = 120 + 15 = 135$ seconds.
From $\displaystyle t = 30$ to $\displaystyle t = 120$ ($30 + 90$), velocity remains constant at $\displaystyle k$.
Let initial acceleration be $\displaystyle a_1 = \displaystyle \frac{k}{30}$.
Deceleration magnitude is $\displaystyle 2a_1 = \displaystyle \frac{2k}{30} = \frac{k}{15}$.
Time taken to come to rest from velocity $\displaystyle k$: $\displaystyle \Delta t = \frac{k}{k/15} = 15$ seconds.
The graph goes down to $\displaystyle v=0$ at $\displaystyle t = 120 + 15 = 135$ seconds.
(a)(ii)
$$ \begin{aligned} \text{Distance during acceleration} &= \frac{1}{2} \times \text{base} \times \text{height} \\ 450 &= \frac{1}{2} \times 30 \times k \\ 15k &= 450 \\ \therefore\quad k &= 30 \text{ ms}^{-1} \\ \text{Acceleration } a &= \frac{k}{30} = \frac{30}{30} = 1 \text{ ms}^{-2} \end{aligned} $$
(b)
$$ \begin{aligned} a &= 3t^2 + 6 \\ v &= \int (3t^2 + 6) \text{ d}t = t^3 + 6t + c \end{aligned} $$
When $\displaystyle t = 0$, $\displaystyle v = 5$:
$$ \begin{aligned} 5 &= (0)^3 + 6(0) + c \implies c = 5 \\ \therefore\quad v &= t^3 + 6t + 5 \end{aligned} $$
When $\displaystyle t = 3$:
$$ \begin{aligned} v &= (3)^3 + 6(3) + 5 \\ v &= 27 + 18 + 5 = 50 \text{ ms}^{-1} \end{aligned} $$
2.
[Source: 0606/12/M/J/15 - Question No. 8]
A particle moves in a straight line such that its displacement, $\displaystyle x$ m, from a fixed point $O$ after $\displaystyle t$ s, is given by $\displaystyle x = 10 \ln(t^{2} + 4) - 4t$.
- Find the initial displacement of the particle from $O$.
- Find the values of $\displaystyle t$ when the particle is instantaneously at rest.
- Find the value of $\displaystyle t$ when the acceleration of the particle is zero.
(a)
Initial displacement is when $\displaystyle t = 0$.
$$ \begin{aligned} x &= 10 \ln(0^2 + 4) - 4(0) \\ x &= 10 \ln 4 \text{ m} \quad (\text{or } 20 \ln 2 \text{ m}) \end{aligned} $$
(b)
$$ \begin{aligned} v = \frac{\text{d}x}{\text{d}t} &= 10\left(\frac{2t}{t^2+4}\right) - 4 \\ &= \frac{20t - 4(t^2+4)}{t^2+4} = \frac{-4t^2 + 20t - 16}{t^2+4} \end{aligned} $$
For instantaneous rest, $\displaystyle v = 0$:
$$ \begin{aligned} -4t^2 + 20t - 16 &= 0 \\ t^2 - 5t + 4 &= 0 \\ (t-1)(t-4) &= 0 \\ \therefore\quad t = 1 \text{ s} \quad &\text{or} \quad t = 4 \text{ s} \end{aligned} $$
(c)
$$ \begin{aligned} a = \frac{\text{d}v}{\text{d}t} &= \frac{\text{d}}{\text{d}t} \left( 20t(t^2+4)^{-1} - 4 \right) \\ &= 20(t^2+4)^{-1} + 20t(-1)(t^2+4)^{-2}(2t) \\ &= \frac{20(t^2+4) - 40t^2}{(t^2+4)^2} = \frac{80 - 20t^2}{(t^2+4)^2} \end{aligned} $$
When acceleration is zero, $\displaystyle a = 0$:
$$ \begin{aligned} 80 - 20t^2 &= 0 \\ 20t^2 &= 80 \implies t^2 = 4 \\ \therefore\quad t &= 2 \text{ s} \quad (\because t \ge 0) \end{aligned} $$
3.
[Source: 0606/21/M/J/15 - Question No. 8]
A particle $P$ is projected from the origin $O$ so that it moves in a straight line. At time $\displaystyle t$ seconds after projection, the velocity of the particle, $\displaystyle v \text{ ms}^{-1}$, is given by $\displaystyle v = 2t^{2} - 14t + 12$.
- Find the time at which $P$ first comes to instantaneous rest.
- Find an expression for the displacement of $P$ from $O$ at time $\displaystyle t$ seconds.
- Find the acceleration of $P$ when $\displaystyle t=3$.
(a)
For instantaneous rest, $\displaystyle v = 0$:
$$ \begin{aligned} 2t^2 - 14t + 12 &= 0 \\ 2(t^2 - 7t + 6) &= 0 \\ 2(t-1)(t-6) &= 0 \\ \therefore\quad t &= 1 \text{ or } t = 6 \end{aligned} $$
$\therefore\quad$ The particle first comes to instantaneous rest at $\displaystyle t = 1$ s.
(b)
$$ \begin{aligned} s &= \int v \text{ d}t = \int (2t^2 - 14t + 12) \text{ d}t \\ s &= \frac{2}{3}t^3 - 7t^2 + 12t + c \end{aligned} $$
Since it is projected from origin $O$, $\displaystyle s = 0$ when $\displaystyle t = 0 \implies c = 0$.
$$ \therefore\quad s = \frac{2}{3}t^3 - 7t^2 + 12t $$
(c)
$$ \begin{aligned} a &= \frac{\text{d}v}{\text{d}t} = 4t - 14 \end{aligned} $$
When $\displaystyle t = 3$:
$$ \begin{aligned} a &= 4(3) - 14 = 12 - 14 = -2 \text{ ms}^{-2} \end{aligned} $$
4.
[Source: 0606/22/O/N/15 - Question No. 10]
A particle is moving in a straight line such that its velocity, $\displaystyle v \text{ ms}^{-1}$, $\displaystyle t$ seconds after passing a fixed point $O$ is $\displaystyle v = e^{2t} - 6e^{-2t} - 1$.
- Find an expression for the displacement, $\displaystyle s$ m, from $O$ of the particle after $\displaystyle t$ seconds.
- Using the substitution $\displaystyle u = e^{2t}$, or otherwise, find the time when the particle is at rest.
- Find the acceleration at this time.
(a)
$$ \begin{aligned} s &= \int v \text{ d}t = \int (e^{2t} - 6e^{-2t} - 1) \text{ d}t \\ s &= \frac{1}{2}e^{2t} + 3e^{-2t} - t + c \end{aligned} $$
Given $\displaystyle s = 0$ when $\displaystyle t = 0$:
$$ \begin{aligned} 0 &= \frac{1}{2}(1) + 3(1) - 0 + c \\ c &= -3.5 \\ \therefore\quad s &= \frac{1}{2}e^{2t} + 3e^{-2t} - t - 3.5 \end{aligned} $$
(b)
For instantaneous rest, $\displaystyle v = 0$:
$$ \begin{aligned} e^{2t} - 6e^{-2t} - 1 &= 0 \end{aligned} $$
Let $\displaystyle u = e^{2t}$, then $\displaystyle e^{-2t} = \displaystyle \frac{1}{u}$:
$$ \begin{aligned} u - \frac{6}{u} - 1 &= 0 \\ u^2 - u - 6 &= 0 \\ (u-3)(u+2) &= 0 \\ \therefore\quad u &= 3 \quad (\text{since } e^{2t} > 0, u = -2 \text{ is rejected}) \\ e^{2t} &= 3 \implies 2t = \ln 3 \implies t = \frac{1}{2}\ln 3 \text{ s} \end{aligned} $$
(c)
$$ \begin{aligned} a &= \frac{\text{d}v}{\text{d}t} = 2e^{2t} + 12e^{-2t} \end{aligned} $$
At $t = \displaystyle \frac{1}{2}\ln 3$, we know $e^{2t} = 3$ and $e^{-2t} = \displaystyle \frac{1}{3}$:
$$ \begin{aligned} a &= 2(3) + 12\left(\frac{1}{3}\right) = 6 + 4 = 10 \text{ ms}^{-2} \end{aligned} $$
5.
[Source: 0606/23/O/N/15 - Question No. 7]
The velocity, $\displaystyle v \text{ ms}^{-1}$, of a particle travelling in a straight line, $\displaystyle t$ seconds after passing through a fixed point $O$, is given by $\displaystyle v = \frac{10}{(2+t)^{2}}$.
- Find the acceleration of the particle when $\displaystyle t = 3$.
- Explain why the particle never comes to rest.
- Find an expression for the displacement of the particle from $O$ after time $\displaystyle t$ s.
- Find the distance travelled by the particle between $\displaystyle t = 3$ and $\displaystyle t = 8$.
(a)
$$ \begin{aligned} v &= 10(2+t)^{-2} \\ a &= \frac{\text{d}v}{\text{d}t} = -20(2+t)^{-3} \end{aligned} $$
When $\displaystyle t = 3$:
$$ \begin{aligned} a &= -20(2+3)^{-3} = -20(125)^{-1} = -\frac{20}{125} = -0.16 \text{ ms}^{-2} \end{aligned} $$
(b)
For all $\displaystyle t \ge 0$, $\displaystyle 10 > 0$ and $\displaystyle (2+t)^2 > 0$.
Therefore, $\displaystyle v = \displaystyle \frac{10}{(2+t)^2} > 0$ for all $\displaystyle t \ge 0$. Since velocity is never zero, the particle never comes to rest.
Therefore, $\displaystyle v = \displaystyle \frac{10}{(2+t)^2} > 0$ for all $\displaystyle t \ge 0$. Since velocity is never zero, the particle never comes to rest.
(c)
$$ \begin{aligned} s &= \int 10(2+t)^{-2} \text{ d}t = \frac{10(2+t)^{-1}}{-1} + c = -\frac{10}{2+t} + c \end{aligned} $$
When $\displaystyle t = 0$, $\displaystyle s = 0$:
$$ \begin{aligned} 0 &= -\frac{10}{2} + c \implies c = 5 \\ \therefore\quad s &= 5 - \frac{10}{2+t} \end{aligned} $$
(d)
Since velocity is always positive, distance is simply the change in displacement.
$$ \begin{aligned} \text{Distance} &= s(8) - s(3) \\ &= \left( 5 - \frac{10}{10} \right) - \left( 5 - \frac{10}{5} \right) \\ &= (5 - 1) - (5 - 2) = 4 - 3 = 1 \text{ m} \end{aligned} $$
6.
[Source: 0606/22/F/M/16 - Question No. 12]
A particle $P$ is projected from the origin $O$ so that it moves in a straight line. At time $\displaystyle t$ seconds after projection, the velocity of the particle, $\displaystyle v \text{ ms}^{-1}$, is given by $\displaystyle v=9t^{2}-63t+90$.
- Show that $P$ first comes to instantaneous rest when $\displaystyle t=2$.
- Find the acceleration of $P$ when $\displaystyle t=3.5$.
- Find an expression for the displacement of $P$ from $O$ at time $\displaystyle t$ seconds.
- Find the distance travelled by $P$
- in the first $2$ seconds,
- in the first $3$ seconds.
(a)
For instantaneous rest, $\displaystyle v = 0$:
$$ \begin{aligned} 9t^2 - 63t + 90 &= 0 \\ 9(t^2 - 7t + 10) &= 0 \\ 9(t-2)(t-5) &= 0 \end{aligned} $$
The times of instantaneous rest are $\displaystyle t=2$ and $\displaystyle t=5$. Thus, $P$ first comes to rest when $\displaystyle t=2$.
(b)
$$ \begin{aligned} a &= \frac{\text{d}v}{\text{d}t} = 18t - 63 \end{aligned} $$
When $\displaystyle t = 3.5$:
$$ \begin{aligned} a &= 18(3.5) - 63 = 63 - 63 = 0 \text{ ms}^{-2} \end{aligned} $$
(c)
$$ \begin{aligned} s &= \int (9t^2 - 63t + 90) \text{ d}t \\ s &= 3t^3 - 31.5t^2 + 90t + c \end{aligned} $$
Projected from origin $\implies \displaystyle s=0$ when $\displaystyle t=0 \implies c=0$.
$$ \therefore\quad s = 3t^3 - 31.5t^2 + 90t $$
(d)(i)
In the first 2 seconds, particle moves continuously in one direction (since $\displaystyle v > 0$).
$$ \begin{aligned} \text{Distance} &= s(2) - s(0) \\ &= 3(2)^3 - 31.5(2)^2 + 90(2) - 0 \\ &= 24 - 126 + 180 = 78 \text{ m} \end{aligned} $$
(d)(ii)
From $\displaystyle t=0$ to $\displaystyle t=2$, particle moves forward. From $\displaystyle t=2$ to $\displaystyle t=3$, velocity is negative (moves backward).
$$ \begin{aligned} s(3) &= 3(3)^3 - 31.5(3)^2 + 90(3) = 81 - 283.5 + 270 = 67.5 \text{ m} \\ \text{Total Distance} &= (\text{Distance } 0 \text{ to } 2) + (\text{Distance } 2 \text{ to } 3) \\ &= 78 + |s(3) - s(2)| \\ &= 78 + |67.5 - 78| = 78 + 10.5 = 88.5 \text{ m} \end{aligned} $$
7.
[Source: 0606/12/M/J/16 - Question No. 11]
- The diagram shows the velocity-time graph of a particle $P$ moving in a straight line with velocity $\displaystyle v \text{ ms}^{-1}$ at time $\displaystyle t$ s after leaving a fixed point.
- Find the distance travelled by the particle $P$.
- Write down the deceleration of the particle when $\displaystyle t=30$.
- The diagram shows a velocity-time graph of a particle $Q$ moving in a straight line with velocity $\displaystyle v \text{ ms}^{-1}$ at time $\displaystyle t$ s after leaving a fixed point.
The displacement of $Q$ at time $\displaystyle t$ s is $\displaystyle s$ m. On the axes below, draw the corresponding displacement-time graph for $Q$.
- The velocity, $\displaystyle v \text{ ms}^{-1}$, of a particle $R$ moving in a straight line, $\displaystyle t$ s after passing through a fixed point $O$, is given by $\displaystyle v=4e^{2t}+6$.
- Explain why the particle is never at rest.
- Find the exact value of $\displaystyle t$ for which the acceleration of $R$ is $\displaystyle 12 \text{ ms}^{-2}$.
- Showing all your working, find the distance travelled by $R$ in the interval between $\displaystyle t=0.4$ and $\displaystyle t=0.5$.
(a)(i)
Distance is the area under the velocity-time graph. The area can be split into a rectangle, a trapezium, another rectangle, and a triangle:
$$ \begin{aligned} \text{Area}_1 (0 \text{ to } 25) &= 25 \times 30 = 750 \\ \text{Area}_2 (25 \text{ to } 35) &= \frac{1}{2}(30+15) \times 10 = 225 \\ \text{Area}_3 (35 \text{ to } 50) &= 15 \times 15 = 225 \\ \text{Area}_4 (50 \text{ to } 60) &= \frac{1}{2} \times 10 \times 15 = 75 \\ \text{Total Distance} &= 750 + 225 + 225 + 75 = 1275 \text{ m} \end{aligned} $$
(a)(ii)
Deceleration at $\displaystyle t=30$ is the negative of the gradient of the line from $\displaystyle t=25$ to $\displaystyle t=35$.
$$ \begin{aligned} \text{Gradient} &= \frac{15 - 30}{35 - 25} = \frac{-15}{10} = -1.5 \\ \therefore\quad \text{Deceleration} &= 1.5 \text{ ms}^{-2} \end{aligned} $$
(b)
For Particle Q:
From $\displaystyle t=0$ to $\displaystyle t=10$, velocity is $6 \text{ ms}^{-1}$. Distance covered = $\displaystyle 10 \times 6 = 60 \text{ m}$.
From $\displaystyle t=10$ to $\displaystyle t=20$, velocity is $3 \text{ ms}^{-1}$. Additional distance = $\displaystyle 10 \times 3 = 30 \text{ m}$.
The displacement-time graph will be a straight line from $(0,0)$ to $(10,60)$, followed by a straight line with a lesser gradient from $(10,60)$ to $(20,90)$.
(c)(i)
$$ v = 4e^{2t} + 6 $$
For any real value of $\displaystyle t$, $\displaystyle e^{2t} > 0$. Therefore, $\displaystyle 4e^{2t} > 0 \implies v > 6$.
Since velocity is always greater than 0, the particle never comes to rest.
(c)(ii)
$$ \begin{aligned} a &= \frac{\text{d}v}{\text{d}t} = 8e^{2t} \\ 12 &= 8e^{2t} \implies e^{2t} = 1.5 \\ \therefore\quad t &= \frac{1}{2} \ln 1.5 \text{ s} \end{aligned} $$
(c)(iii)
$$ \begin{aligned} \text{Distance} &= \int_{0.4}^{0.5} (4e^{2t} + 6) \text{ d}t \\ &= \left[ 2e^{2t} + 6t \right]_{0.4}^{0.5} \\ &= \left( 2e^{2(0.5)} + 6(0.5) \right) - \left( 2e^{2(0.4)} + 6(0.4) \right) \\ &= (2e^1 + 3) - (2e^{0.8} + 2.4) \\ &= 2e - 2e^{0.8} + 0.6 \text{ m} \end{aligned} $$
8.
[Source: 0606/12/M/J/17 - Question No. 5]
A particle $P$ moves in a straight line, such that its displacement, $\displaystyle x$ m, from a fixed point $O$, $\displaystyle t$ s after passing $O$, is given by $\displaystyle x = 4 \cos(3t) - 4$.
- Find the velocity of $P$ at time $\displaystyle t$.
- Hence write down the maximum speed of $P$.
- Find the smallest value of $\displaystyle t$ for which the acceleration of $P$ is zero.
- For the value of $\displaystyle t$ found in part (iii), find the distance of $P$ from $O$.
(a)
$$ v = \frac{\text{d}x}{\text{d}t} = -12 \sin(3t) \text{ ms}^{-1} $$
(b)
The maximum value of $\displaystyle |-\sin(3t)|$ is 1.
$$ \therefore\quad \text{Maximum speed} = 12 \text{ ms}^{-1} $$
(c)
$$ \begin{aligned} a &= \frac{\text{d}v}{\text{d}t} = -36 \cos(3t) \\ 0 &= -36 \cos(3t) \implies \cos(3t) = 0 \end{aligned} $$
The smallest positive value for $\displaystyle 3t$ is $\displaystyle \frac{\pi}{2}$:
$$ 3t = \frac{\pi}{2} \implies t = \frac{\pi}{6} \text{ s} $$
(d)
Substitute $\displaystyle t = \displaystyle \frac{\pi}{6}$ into the displacement equation:
$$ \begin{aligned} x &= 4 \cos\left(3\left(\frac{\pi}{6}\right)\right) - 4 \\ x &= 4 \cos\left(\frac{\pi}{2}\right) - 4 = 4(0) - 4 = -4 \text{ m} \end{aligned} $$
Distance is the magnitude of displacement, therefore distance from $O$ is $4 \text{ m}$.
9.
[Source: 0606/13/M/J/17 - Question No. 12]
A particle moves in a straight line, such that its velocity, $\displaystyle v \text{ ms}^{-1}$, $\displaystyle t$ s after passing a fixed point $O$, is given by $\displaystyle v = 2 + 6t + 3 \sin 2t$.
- Find the acceleration of the particle at time $\displaystyle t$.
- Hence find the smallest value of $\displaystyle t$ for which the acceleration of the particle is zero.
- Find the displacement, $\displaystyle x$ m from $O$, of the particle at time $\displaystyle t$.
(a)
$$ a = \frac{\text{d}v}{\text{d}t} = 6 + 6 \cos(2t) \text{ ms}^{-2} $$
(b)
For zero acceleration:
$$ \begin{aligned} 6 + 6 \cos(2t) &= 0 \\ \cos(2t) &= -1 \end{aligned} $$
Smallest positive value of $\displaystyle t$:
$$ 2t = \pi \implies t = \frac{\pi}{2} \text{ s} $$
(c)
$$ \begin{aligned} x &= \int (2 + 6t + 3 \sin 2t) \text{ d}t \\ x &= 2t + 3t^2 - \frac{3}{2} \cos(2t) + c \end{aligned} $$
Since $\displaystyle x=0$ when $\displaystyle t=0$:
$$ \begin{aligned} 0 &= 2(0) + 3(0) - 1.5(1) + c \implies c = 1.5 \\ \therefore\quad x &= 3t^2 + 2t - 1.5 \cos(2t) + 1.5 \end{aligned} $$
10.
[Source: 0606/21/M/J/17 - Question No. 12]
A particle moves in a straight line so that, $\displaystyle t$ seconds after passing a fixed point $O$, its displacement, $\displaystyle s$ m, from $O$ is given by $\displaystyle s = 1 + 3t - \cos 5t$.
- Find the distance between the particle's first two positions of instantaneous rest.
- Find the acceleration when $\displaystyle t = \pi$.
(a)
$$ v = \frac{\text{d}s}{\text{d}t} = 3 + 5 \sin(5t) $$
For instantaneous rest, $\displaystyle v = 0$:
$$ \begin{aligned} 3 + 5 \sin(5t) &= 0 \implies \sin(5t) = -0.6 \end{aligned} $$
The first two positive solutions for $\displaystyle 5t$ are in the 3rd and 4th quadrants. Let $\displaystyle \alpha = \arcsin(0.6)$.
$$ 5t_1 = \pi + \alpha, \quad 5t_2 = 2\pi - \alpha $$
Since velocity stays negative between these two times, the particle moves in one direction without reversing. Distance is just the difference in displacement:
$$ \text{Distance} = |s(t_2) - s(t_1)| $$
Substitute $t_1, t_2$ back to displacement equation. Note that $\displaystyle \cos(5t_1) = -\sqrt{1 - (-0.6)^2} = -0.8$ and $\displaystyle \cos(5t_2) = +0.8$.
$$ \begin{aligned} s(t_1) &= 1 + \frac{3}{5}(\pi + \alpha) - (-0.8) = 1.8 + 0.6\pi + 0.6\alpha \\ s(t_2) &= 1 + \frac{3}{5}(2\pi - \alpha) - (+0.8) = 0.2 + 1.2\pi - 0.6\alpha \end{aligned} $$
$$ \begin{aligned} \text{Distance} &= |(0.2 + 1.2\pi - 0.6\alpha) - (1.8 + 0.6\pi + 0.6\alpha)| \\ &= |-1.6 + 0.6\pi - 1.2\alpha| \\ &= 1.6 - 0.6\pi + 1.2\arcsin(0.6) \\ &\approx 1.6 - 1.885 + 0.772 = 0.487 \text{ m} \end{aligned} $$
(b)
$$ \begin{aligned} a &= \frac{\text{d}v}{\text{d}t} = 25 \cos(5t) \end{aligned} $$
When $\displaystyle t = \pi$:
$$ a = 25 \cos(5\pi) = 25(-1) = -25 \text{ ms}^{-2} $$
11.
[Source: 0606/12/O/N/17 - Question No. 9]
- The diagram shows the displacement-time graph of a particle $P$ which moves in a straight line such that, $\displaystyle t$ s after leaving a fixed point $O$, its displacement from $O$ is $\displaystyle x$ m.
On the axes below, draw the velocity-time graph of $P$.
- A particle $Q$ moves in a straight line such that its velocity, $\displaystyle v \text{ ms}^{-1}$, $\displaystyle t$ s after passing through a fixed point $O$, is given by $\displaystyle v = 3e^{-5t} + \frac{3t}{2}$, for $\displaystyle t \ge 0$.
- Find the velocity of $Q$ when $\displaystyle t = 0$.
- Find the value of $\displaystyle t$ when the acceleration of $Q$ is zero.
- Find the distance of $Q$ from $O$ when $\displaystyle t = 0.5$.
(a)
Velocity is the gradient of the displacement-time graph.
From $\displaystyle t = 0$ to $\displaystyle t = 10$, $\displaystyle v = \frac{60 - 0}{10 - 0} = 6 \text{ ms}^{-1}$.
From $\displaystyle t = 10$ to $\displaystyle t = 20$, $\displaystyle v = \frac{60 - 60}{20 - 10} = 0 \text{ ms}^{-1}$.
From $\displaystyle t = 20$ to $\displaystyle t = 30$, $\displaystyle v = \frac{0 - 60}{30 - 20} = -6 \text{ ms}^{-1}$.
From $\displaystyle t = 0$ to $\displaystyle t = 10$, $\displaystyle v = \frac{60 - 0}{10 - 0} = 6 \text{ ms}^{-1}$.
From $\displaystyle t = 10$ to $\displaystyle t = 20$, $\displaystyle v = \frac{60 - 60}{20 - 10} = 0 \text{ ms}^{-1}$.
From $\displaystyle t = 20$ to $\displaystyle t = 30$, $\displaystyle v = \frac{0 - 60}{30 - 20} = -6 \text{ ms}^{-1}$.
(b)(i)
$$ \begin{aligned} v(0) &= 3e^{-5(0)} + \frac{3(0)}{2} \\ &= 3(1) + 0 = 3 \text{ ms}^{-1} \end{aligned} $$
(b)(ii)
$$ \begin{aligned} a &= \frac{\text{d}v}{\text{d}t} = -15e^{-5t} + 1.5 \\ 0 &= -15e^{-5t} + 1.5 \\ 15e^{-5t} &= 1.5 \\ e^{-5t} &= 0.1 \\ -5t &= \ln(0.1) \\ \therefore\quad t &= -0.2 \ln(0.1) \quad (\text{or } 0.2 \ln 10) \text{ s} \end{aligned} $$
(b)(iii)
$$ \begin{aligned} s &= \int \left( 3e^{-5t} + 1.5t \right) \text{ d}t \\ s &= -\frac{3}{5}e^{-5t} + \frac{1.5t^2}{2} + c = -0.6e^{-5t} + 0.75t^2 + c \end{aligned} $$
When $\displaystyle t = 0$, $\displaystyle s = 0$:
$$ \begin{aligned} 0 &= -0.6(1) + 0 + c \implies c = 0.6 \\ \therefore\quad s &= 0.6 - 0.6e^{-5t} + 0.75t^2 \end{aligned} $$
When $\displaystyle t = 0.5$:
$$ \begin{aligned} s &= 0.6 - 0.6e^{-5(0.5)} + 0.75(0.5)^2 \\ &= 0.6 - 0.6e^{-2.5} + 0.75(0.25) \\ &= 0.6 + 0.1875 - 0.6e^{-2.5} \\ &= 0.7875 - 0.6e^{-2.5} \text{ m} \end{aligned} $$
12.
[Source: 0606/12/F/M/18 - Question No. 8]
A particle $P$, moving in a straight line, passes through a fixed point $O$ at time $\displaystyle t = 0$ s. At time $\displaystyle t$ s after leaving $O$, the displacement of the particle is $\displaystyle x$ m and its velocity is $\displaystyle v \text{ ms}^{-1}$, where $\displaystyle v = 12e^{2t} - 48t$, for $\displaystyle t \ge 0$.
- Find $\displaystyle x$ in terms of $\displaystyle t$.
- Find the value of $\displaystyle t$ when the acceleration of $P$ is zero.
- Find the velocity of $P$ when the acceleration is zero.
(a)
$$ \begin{aligned} x &= \int v \text{ d}t = \int (12e^{2t} - 48t) \text{ d}t \\ x &= 6e^{2t} - 24t^2 + c \end{aligned} $$
Since $\displaystyle x=0$ when $\displaystyle t=0$:
$$ \begin{aligned} 0 &= 6e^0 - 24(0) + c \implies c = -6 \\ \therefore\quad x &= 6e^{2t} - 24t^2 - 6 \end{aligned} $$
(b)
$$ \begin{aligned} a &= \frac{\text{d}v}{\text{d}t} = 24e^{2t} - 48 \end{aligned} $$
When $\displaystyle a = 0$:
$$ \begin{aligned} 24e^{2t} - 48 &= 0 \\ 24e^{2t} &= 48 \implies e^{2t} = 2 \\ 2t &= \ln 2 \\ \therefore\quad t &= \frac{1}{2}\ln 2 \text{ s} \end{aligned} $$
(c)
Substitute $\displaystyle t = \frac{1}{2}\ln 2$ into $\displaystyle v$:
$$ \begin{aligned} v &= 12e^{2(0.5\ln 2)} - 48(0.5\ln 2) \\ v &= 12(2) - 24\ln 2 \\ \therefore\quad v &= 24 - 24\ln 2 \text{ ms}^{-1} \end{aligned} $$
13.
[Source: 0606/12/M/J/18 - Question No. 4]
A particle $P$ moves so that its displacement, $\displaystyle x$ metres from a fixed point $O$, at time $\displaystyle t$ seconds, is given by $\displaystyle x = \ln(5t + 3)$.
- Find the value of $\displaystyle t$ when the displacement of $P$ is $3$ m.
- Find the velocity of $P$ when $\displaystyle t = 0$.
- Explain why, after passing through $O$, the velocity of $P$ is never negative.
- Find the acceleration of $P$ when $\displaystyle t = 0$.
(a)
$$ \begin{aligned} \ln(5t + 3) &= 3 \\ 5t + 3 &= e^3 \\ 5t &= e^3 - 3 \\ \therefore\quad t &= \frac{e^3 - 3}{5} \text{ s} \end{aligned} $$
(b)
$$ \begin{aligned} v &= \frac{\text{d}x}{\text{d}t} = \frac{5}{5t+3} \end{aligned} $$
When $\displaystyle t = 0$:
$$ \begin{aligned} v &= \frac{5}{5(0)+3} = \frac{5}{3} \text{ ms}^{-1} \end{aligned} $$
(c)
Since $\displaystyle t \ge 0$, the expression $\displaystyle (5t+3)$ is always positive ($\displaystyle 5t+3 \ge 3 > 0$).
Therefore, $\displaystyle v = \frac{5}{5t+3}$ is a positive value divided by a positive value, which makes $\displaystyle v > 0$ for all $\displaystyle t \ge 0$. So it is never negative.
Therefore, $\displaystyle v = \frac{5}{5t+3}$ is a positive value divided by a positive value, which makes $\displaystyle v > 0$ for all $\displaystyle t \ge 0$. So it is never negative.
(d)
$$ \begin{aligned} a &= \frac{\text{d}v}{\text{d}t} = \frac{\text{d}}{\text{d}t}\left( 5(5t+3)^{-1} \right) \\ a &= -5(5t+3)^{-2}(5) = \frac{-25}{(5t+3)^2} \end{aligned} $$
When $\displaystyle t = 0$:
$$ \begin{aligned} a &= \frac{-25}{(3)^2} = -\frac{25}{9} \text{ ms}^{-2} \end{aligned} $$
14.
[Source: 0606/21/M/J/18 - Question No. 10]
A particle moves in a straight line such that its displacement, $\displaystyle s$ metres, from a fixed point $O$ at time $\displaystyle t$ seconds, is given by $\displaystyle s = 4 + \cos 3t$, where $\displaystyle t \ge 0$. The particle is initially at rest.
- Find the exact value of $\displaystyle t$ when the particle is next at rest.
- Find the distance travelled by the particle between $\displaystyle t = \frac{\pi}{4}$ and $\displaystyle t = \frac{\pi}{2}$ seconds.
- Find the greatest acceleration of the particle.
(a)
$$ \begin{aligned} v &= \frac{\text{d}s}{\text{d}t} = -3\sin(3t) \end{aligned} $$
For rest, $\displaystyle v = 0$:
$$ \begin{aligned} -3\sin(3t) &= 0 \implies \sin(3t) = 0 \end{aligned} $$
The next time it is at rest is when $\displaystyle 3t = \pi$:
$$ \therefore\quad t = \frac{\pi}{3} \text{ s} $$
(b)
The particle changes direction at $\displaystyle t = \frac{\pi}{3}$, which falls between $\displaystyle t = \frac{\pi}{4}$ and $\displaystyle t = \frac{\pi}{2}$.
$$ \begin{aligned} s\left(\frac{\pi}{4}\right) &= 4 + \cos\left(\frac{3\pi}{4}\right) = 4 - \frac{\sqrt{2}}{2} \\ s\left(\frac{\pi}{3}\right) &= 4 + \cos(\pi) = 4 - 1 = 3 \\ s\left(\frac{\pi}{2}\right) &= 4 + \cos\left(\frac{3\pi}{2}\right) = 4 + 0 = 4 \end{aligned} $$
$$ \begin{aligned} \text{Distance} &= \left| s\left(\frac{\pi}{3}\right) - s\left(\frac{\pi}{4}\right) \right| + \left| s\left(\frac{\pi}{2}\right) - s\left(\frac{\pi}{3}\right) \right| \\ &= \left| 3 - \left( 4 - \frac{\sqrt{2}}{2} \right) \right| + \left| 4 - 3 \right| \\ &= \left| \frac{\sqrt{2}}{2} - 1 \right| + 1 \\ &= 1 - \frac{\sqrt{2}}{2} + 1 = 2 - \frac{\sqrt{2}}{2} \text{ m} \end{aligned} $$
(c)
$$ \begin{aligned} a &= \frac{\text{d}v}{\text{d}t} = -9\cos(3t) \end{aligned} $$
The maximum value of $\displaystyle \cos(3t)$ is $1$ and minimum is $-1$.
$$ \therefore\quad \text{Greatest acceleration} = -9(-1) = 9 \text{ ms}^{-2} $$
15.
[Source: 0606/12/F/M/19 - Question No. 8]
- The diagram shows the velocity-time graph of a particle $P$ moving in a straight line with velocity $\displaystyle v \text{ ms}^{-1}$ at time $\displaystyle t$ seconds after leaving a fixed point.
- Write down the value of the acceleration of $P$ when $\displaystyle t = 5$.
- Find the distance travelled by the particle $P$ between $\displaystyle t = 0$ and $\displaystyle t = 10$.
- A particle $Q$ moves such that its velocity, $\displaystyle v \text{ ms}^{-1}$, $\displaystyle t$ seconds after leaving a fixed point, is given by $\displaystyle v = 3 \sin 2t - 1$.
- Find the speed of $Q$ when $\displaystyle t = \frac{7\pi}{12}$.
- Find the least value of $\displaystyle t$ for which the acceleration of $Q$ is zero.
(a)(i)
At $\displaystyle t = 5$, the graph is a horizontal line (from $t=2$ to $t=6$).
Therefore, the gradient is 0. $$ \therefore\quad \text{Acceleration } a = 0 \text{ ms}^{-2} $$
Therefore, the gradient is 0. $$ \therefore\quad \text{Acceleration } a = 0 \text{ ms}^{-2} $$
(a)(ii)
Distance is the area under the v-t graph.
$$ \begin{aligned} \text{Area}_1 (0 \text{ to } 2) &= \frac{1}{2} \times 2 \times 10 = 10 \\ \text{Area}_2 (2 \text{ to } 6) &= 4 \times 10 = 40 \\ \text{Area}_3 (6 \text{ to } 10) &= \frac{1}{2} \times (10 + 20) \times 4 = 60 \\ \text{Total Distance} &= 10 + 40 + 60 = 110 \text{ m} \end{aligned} $$
$$ \begin{aligned} \text{Area}_1 (0 \text{ to } 2) &= \frac{1}{2} \times 2 \times 10 = 10 \\ \text{Area}_2 (2 \text{ to } 6) &= 4 \times 10 = 40 \\ \text{Area}_3 (6 \text{ to } 10) &= \frac{1}{2} \times (10 + 20) \times 4 = 60 \\ \text{Total Distance} &= 10 + 40 + 60 = 110 \text{ m} \end{aligned} $$
(b)(i)
$$ \begin{aligned} v &= 3 \sin\left(2\left(\frac{7\pi}{12}\right)\right) - 1 \\ &= 3 \sin\left(\frac{7\pi}{6}\right) - 1 \\ &= 3(-0.5) - 1 = -1.5 - 1 = -2.5 \text{ ms}^{-1} \end{aligned} $$
Speed is the magnitude of velocity.
$$ \therefore\quad \text{Speed} = |-2.5| = 2.5 \text{ ms}^{-1} $$
(b)(ii)
$$ \begin{aligned} a &= \frac{\text{d}v}{\text{d}t} = 6 \cos(2t) \end{aligned} $$
For zero acceleration:
$$ \begin{aligned} 6 \cos(2t) &= 0 \implies \cos(2t) = 0 \end{aligned} $$
Least positive value:
$$ 2t = \frac{\pi}{2} \implies t = \frac{\pi}{4} \text{ s} $$
16.
[Source: 0606/21/M/J/19 - Question No. 5]
The velocity-time graph represents the motion of a particle travelling in a straight line.
- Find the acceleration during the last 6 seconds of the motion.
- The particle travels with constant velocity for 23 seconds. Find the value of $\displaystyle k$.
- Using your answer to part (b), find the total distance travelled by the particle.
(a)
The last 6 seconds correspond to the interval from $\displaystyle t = k$ to $\displaystyle t = k+6$.
The velocity goes from $10 \text{ ms}^{-1}$ down to $0 \text{ ms}^{-1}$.
$$ \begin{aligned} a &= \text{gradient} = \frac{0 - 10}{6} = -\frac{10}{6} = -\frac{5}{3} \text{ ms}^{-2} \end{aligned} $$
(b)
The period of constant velocity is from $\displaystyle t = 4$ to $\displaystyle t = k$.
$$ \begin{aligned} k - 4 &= 23 \\ \therefore\quad k &= 27 \end{aligned} $$
(c)
The shape is a trapezium.
$$ \begin{aligned} \text{Parallel sides: } & a = k - 4 = 23 \\ & b = k + 6 = 27 + 6 = 33 \\ \text{Height: } & h = 10 \\ \text{Distance} &= \frac{1}{2}(a+b)h = \frac{1}{2}(23 + 33) \times 10 \\ &= \frac{1}{2}(56) \times 10 = 280 \text{ m} \end{aligned} $$
17.
[Source: 0606/22/M/J/19 - Question No. 11]
The velocity, $\displaystyle v \text{ ms}^{-1}$, of a particle travelling in a straight line, $\displaystyle t$ seconds after passing through a fixed point $O$, is given by $\displaystyle v=\frac{4}{(t+1)^{3}}$.
- Explain why the direction of motion of the particle never changes.
- Showing all your working, find the acceleration of the particle when $\displaystyle t=5$.
- Find an expression for the displacement of the particle from $O$ after $\displaystyle t$ seconds.
- Find the distance travelled by the particle in the fourth second.
(a)
For all $\displaystyle t \ge 0$, $\displaystyle (t+1) > 0$, so $\displaystyle (t+1)^3 > 0$.
Therefore, $\displaystyle v = \frac{4}{(t+1)^3} > 0$. Since the velocity is always positive and never changes sign, the direction of motion never changes.
Therefore, $\displaystyle v = \frac{4}{(t+1)^3} > 0$. Since the velocity is always positive and never changes sign, the direction of motion never changes.
(b)
$$ \begin{aligned} a &= \frac{\text{d}v}{\text{d}t} = 4(-3)(t+1)^{-4} = \frac{-12}{(t+1)^4} \end{aligned} $$
When $\displaystyle t = 5$:
$$ \begin{aligned} a &= \frac{-12}{(5+1)^4} = \frac{-12}{6^4} = \frac{-12}{1296} = -\frac{1}{108} \text{ ms}^{-2} \end{aligned} $$
(c)
$$ \begin{aligned} s &= \int 4(t+1)^{-3} \text{ d}t = \frac{4(t+1)^{-2}}{-2} + c = -\frac{2}{(t+1)^2} + c \end{aligned} $$
Given $\displaystyle s = 0$ when $\displaystyle t = 0$:
$$ \begin{aligned} 0 &= -2(1)^{-2} + c \implies c = 2 \\ \therefore\quad s &= 2 - \frac{2}{(t+1)^2} \end{aligned} $$
(d)
The fourth second is the interval between $\displaystyle t = 3$ and $\displaystyle t = 4$.
$$ \begin{aligned} \text{Distance} &= s(4) - s(3) \\ &= \left( 2 - \frac{2}{5^2} \right) - \left( 2 - \frac{2}{4^2} \right) \\ &= -\frac{2}{25} + \frac{2}{16} = \frac{1}{8} - \frac{2}{25} \\ &= \frac{25 - 16}{200} = \frac{9}{200} \text{ m} \quad (\text{or } 0.045 \text{ m}) \end{aligned} $$
18.
[Source: 0606/23/M/J/19 - Question No. 11]
A particle travelling in a straight line passes through a fixed point $O$. The displacement, $\displaystyle x$ metres, of the particle, $\displaystyle t$ seconds after it passes through $O$, is given by $\displaystyle x=5t+\sin t$.
- Show that the particle is never at rest.
- Find the distance travelled by the particle between $\displaystyle t=\frac{\pi}{3}$ and $\displaystyle t=\frac{\pi}{2}$.
- Find the acceleration of the particle when $\displaystyle t=4$.
- Find the value of $\displaystyle t$ when the velocity of the particle is first at its minimum.
(a)
$$ \begin{aligned} v &= \frac{\text{d}x}{\text{d}t} = 5 + \cos t \end{aligned} $$
Since the range of $\displaystyle \cos t$ is $\displaystyle -1 \le \cos t \le 1$,
$$ v \ge 5 - 1 = 4 $$
Since $\displaystyle v \ge 4$, velocity is always strictly greater than zero. Thus, the particle is never at rest.
(b)
Since the particle only moves in one direction, distance is the difference in displacement.
$$ \begin{aligned} x\left(\frac{\pi}{2}\right) &= 5\left(\frac{\pi}{2}\right) + \sin\left(\frac{\pi}{2}\right) = \frac{5\pi}{2} + 1 \\ x\left(\frac{\pi}{3}\right) &= 5\left(\frac{\pi}{3}\right) + \sin\left(\frac{\pi}{3}\right) = \frac{5\pi}{3} + \frac{\sqrt{3}}{2} \\ \text{Distance} &= x\left(\frac{\pi}{2}\right) - x\left(\frac{\pi}{3}\right) \\ &= \frac{5\pi}{2} - \frac{5\pi}{3} + 1 - \frac{\sqrt{3}}{2} \\ &= \frac{5\pi}{6} + 1 - \frac{\sqrt{3}}{2} \text{ m} \end{aligned} $$
(c)
$$ \begin{aligned} a &= \frac{\text{d}v}{\text{d}t} = -\sin t \end{aligned} $$
When $\displaystyle t = 4$:
$$ \begin{aligned} a &= -\sin 4 \text{ ms}^{-2} \quad (\approx 0.757 \text{ ms}^{-2}) \end{aligned} $$
(d)
Velocity $\displaystyle v = 5 + \cos t$.
The velocity is minimum when $\displaystyle \cos t$ is minimum, which is $\displaystyle -1$.
$$ \begin{aligned} \cos t &= -1 \end{aligned} $$
The first positive value of $\displaystyle t$ is:
$$ t = \pi \text{ s} $$
19.
[Source: 0606/13/O/N/19 - Question No. 10]
- The velocity-time graph for a particle $P$ is shown by the two straight lines in the diagram.
- Find the deceleration of $P$ for $\displaystyle 5 \le t \le 10$.
- Write down the value of $\displaystyle t$ when the speed of $P$ is zero.
- Find the distance $P$ has travelled for $\displaystyle 0 \le t \le 10$.
- A particle $Q$ has a displacement of $\displaystyle x$ m from a fixed point $O$, $\displaystyle t$ s after leaving $O$. The velocity, $\displaystyle v \text{ ms}^{-1}$, of $Q$ at time $\displaystyle t$ s is given by $\displaystyle v = 6e^{2t} + 1$.
- Find an expression for $\displaystyle x$ in terms of $\displaystyle t$.
- Find the value of $\displaystyle t$ when the acceleration of $Q$ is $\displaystyle 24 \text{ ms}^{-2}$.
(a)(i)
For $\displaystyle 5 \le t \le 10$, the velocity goes from $0$ to $-20$.
$$ \begin{aligned} a &= \text{gradient} = \frac{-20 - 0}{10 - 5} = \frac{-20}{5} = -4 \text{ ms}^{-2} \\ \therefore\quad \text{Deceleration} &= 4 \text{ ms}^{-2} \end{aligned} $$
(a)(ii)
Speed is zero when the graph crosses the t-axis.
$$ \therefore\quad t = 5 \text{ s} $$
(a)(iii)
Total distance is the sum of the areas of the two triangles (without negative signs).
$$ \begin{aligned} \text{Area 1 } (0 \to 5) &= \frac{1}{2} \times 5 \times 20 = 50 \\ \text{Area 2 } (5 \to 10) &= \frac{1}{2} \times 5 \times |-20| = 50 \\ \text{Total Distance} &= 50 + 50 = 100 \text{ m} \end{aligned} $$
(b)(i)
$$ \begin{aligned} x &= \int (6e^{2t} + 1) \text{ d}t = 3e^{2t} + t + c \end{aligned} $$
Leaving $O$ implies $\displaystyle x = 0$ when $\displaystyle t = 0$:
$$ \begin{aligned} 0 &= 3e^0 + 0 + c \implies c = -3 \\ \therefore\quad x &= 3e^{2t} + t - 3 \end{aligned} $$
(b)(ii)
$$ \begin{aligned} a &= \frac{\text{d}v}{\text{d}t} = 12e^{2t} \end{aligned} $$
When $\displaystyle a = 24$:
$$ \begin{aligned} 12e^{2t} &= 24 \\ e^{2t} &= 2 \\ 2t &= \ln 2 \\ \therefore\quad t &= 0.5 \ln 2 \text{ s} \quad (\text{or } \ln\sqrt{2}) \end{aligned} $$
20.
[Source: 0606/21/O/N/19 - Question No. 5]
A particle is moving in a straight line such that $\displaystyle t$ seconds after passing a fixed point $O$ its displacement, $\displaystyle s$ m, is given by $\displaystyle s = 3 \sin 2t + 4 \cos 2t - 4$.
- Find expressions for the velocity and acceleration of the particle at time $\displaystyle t$.
- Find the first time when the particle is instantaneously at rest.
- Find the acceleration of the particle at the time found in part (b).
(a)
$$ \begin{aligned} v &= \frac{\text{d}s}{\text{d}t} = 6 \cos 2t - 8 \sin 2t \\ a &= \frac{\text{d}v}{\text{d}t} = -12 \sin 2t - 16 \cos 2t \end{aligned} $$
(b)
For instantaneous rest, $\displaystyle v = 0$:
$$ \begin{aligned} 6 \cos 2t - 8 \sin 2t &= 0 \\ 8 \sin 2t &= 6 \cos 2t \\ \tan 2t &= \frac{6}{8} = 0.75 \end{aligned} $$
The first positive solution:
$$ \begin{aligned} 2t &= \arctan(0.75) \approx 0.6435 \\ \therefore\quad t &\approx 0.322 \text{ s} \end{aligned} $$
(c)
From $\displaystyle \tan 2t = \frac{3}{4}$ (in the first quadrant), we can deduce the exact values:
$$ \sin 2t = \frac{3}{5} \quad \text{and} \quad \cos 2t = \frac{4}{5} $$
Substitute into acceleration equation:
$$ \begin{aligned} a &= -12\left(\frac{3}{5}\right) - 16\left(\frac{4}{5}\right) \\ &= -\frac{36}{5} - \frac{64}{5} \\ &= -\frac{100}{5} = -20 \text{ ms}^{-2} \end{aligned} $$
21.
[Source: 0606/22/F/M/20 - Question No. 12]
A particle $P$ moves in a straight line such that, $t$ seconds after passing through a fixed point $O$, its acceleration, $a \text{ ms}^{-2}$, is given by $a = -6$. When $t = 0$, the velocity of $P$ is $18 \text{ ms}^{-1}$.
- Find the time at which $P$ comes to instantaneous rest.
- Find the distance travelled by $P$ in the 3rd second.
(a)
$$ \begin{aligned} a &= -6 \\ v &= \int -6 \text{ d}t = -6t + c \end{aligned} $$
When $t = 0$, $v = 18$:
$$ \begin{aligned} 18 &= -6(0) + c \implies c = 18 \\ \therefore\quad v &= 18 - 6t \end{aligned} $$
For instantaneous rest, $v = 0$:
$$ \begin{aligned} 18 - 6t &= 0 \\ 6t &= 18 \\ \therefore\quad t &= 3 \text{ s} \end{aligned} $$
(b)
The 3rd second is the interval between $t = 2$ and $t = 3$.
Since the particle comes to rest at $t = 3$, it does not change direction during this interval.
$$ \begin{aligned} s &= \int (18 - 6t) \text{ d}t = 18t - 3t^2 + d \end{aligned} $$
Distance is $|s(3) - s(2)|$:
$$ \begin{aligned} s(3) &= 18(3) - 3(3^2) + d = 54 - 27 + d = 27 + d \\ s(2) &= 18(2) - 3(2^2) + d = 36 - 12 + d = 24 + d \\ \text{Distance} &= |(27 + d) - (24 + d)| = 3 \text{ m} \end{aligned} $$
22.
[Source: 0606/13/M/J/20 - Question No. 5]
At time $t$ s, a particle travelling in a straight line has acceleration $(2t+1)^{-\frac{1}{2}} \text{ ms}^{-2}$. When $t=0$, the particle is $4$ m from a fixed point $O$ and is travelling with velocity $8 \text{ ms}^{-1}$ away from $O$.
- Find the velocity of the particle at time $t$ s.
- Find the displacement of the particle from $O$ at time $t$ s.
(a)
$$ \begin{aligned} a &= (2t+1)^{-1/2} \\ v &= \int (2t+1)^{-1/2} \text{ d}t = \frac{(2t+1)^{1/2}}{1/2 \times 2} + c = (2t+1)^{1/2} + c \end{aligned} $$
When $t = 0$, $v = 8$:
$$ \begin{aligned} 8 &= (1)^{1/2} + c \implies c = 7 \\ \therefore\quad v &= (2t+1)^{1/2} + 7 \end{aligned} $$
(b)
$$ \begin{aligned} s &= \int \left( (2t+1)^{1/2} + 7 \right) \text{ d}t \\ s &= \frac{(2t+1)^{3/2}}{3/2 \times 2} + 7t + d = \frac{1}{3}(2t+1)^{3/2} + 7t + d \end{aligned} $$
When $t = 0$, $s = 4$:
$$ \begin{aligned} 4 &= \frac{1}{3}(1)^{3/2} + 0 + d \implies 4 = \frac{1}{3} + d \implies d = \frac{11}{3} \\ \therefore\quad s &= \frac{1}{3}(2t+1)^{3/2} + 7t + \frac{11}{3} \end{aligned} $$
23.
[Source: 0606/21/M/J/20 - Question No. 9]
A particle travels in a straight line. As it passes through a fixed point $O$, the particle is travelling at a velocity of $3 \text{ ms}^{-1}$. The particle continues at this velocity for $60$ seconds then decelerates at a constant rate for $15$ seconds to a velocity of $1.6 \text{ ms}^{-1}$. The particle then decelerates again at a constant rate for $5$ seconds to reach point $A$, where it stops.
- Sketch the velocity-time graph for this journey on the axes below.
- Find the distance between $O$ and $A$.
- Find the deceleration in the last $5$ seconds.
(b)
Distance is the total area under the velocity-time graph.
$$ \begin{aligned} \text{Area 1 (Rectangle)} &= 60 \times 3 = 180 \\ \text{Area 2 (Trapezium)} &= \frac{1}{2} (3 + 1.6) \times 15 = \frac{1}{2} (4.6) \times 15 = 2.3 \times 15 = 34.5 \\ \text{Area 3 (Triangle)} &= \frac{1}{2} \times 5 \times 1.6 = 4 \\ \text{Total Distance} &= 180 + 34.5 + 4 = 218.5 \text{ m} \end{aligned} $$
(c)
The last 5 seconds are from $t=75$ to $t=80$, where velocity goes from $1.6 \text{ ms}^{-1}$ to $0$.
$$ \begin{aligned} \text{Acceleration} &= \frac{0 - 1.6}{5} = -0.32 \text{ ms}^{-2} \\ \therefore\quad \text{Deceleration} &= 0.32 \text{ ms}^{-2} \end{aligned} $$
24.
[Source: 0606/12/O/N/20 - Question No. 12]
- The diagram shows the velocity-time graph of a particle $P$ that travels $2775$ m in $90$ s, reaching a final velocity of $V \text{ ms}^{-1}$.
- Find the value of $V$.
- Write down the acceleration of $P$ when $t = 40$.
- The acceleration, $a \text{ ms}^{-2}$, of a particle $Q$ travelling in a straight line, is given by $a = 6 \cos 2t$ at time $t$ s. When $t = 0$ the particle is at point $O$ and is travelling with a velocity of $10 \text{ ms}^{-1}$.
- Find the velocity of $Q$ at time $t$.
- Find the displacement of $Q$ from $O$ at time $t$.
(a)(i)
Total distance = Area under the graph = 2775.
$$ \begin{aligned} \text{Area}_{0-20} &= \frac{1}{2} \times 20 \times 30 = 300 \\ \text{Area}_{20-60} &= 40 \times 30 = 1200 \\ \text{Area}_{60-90} &= \frac{1}{2}(30 + V) \times 30 = 15(30 + V) = 450 + 15V \\ \text{Total} &= 300 + 1200 + 450 + 15V = 1950 + 15V \\ 1950 + 15V &= 2775 \\ 15V &= 825 \\ \therefore\quad V &= 55 \end{aligned} $$
(a)(ii)
At $t = 40$, the graph is horizontal (constant velocity).
$$ \therefore\quad a = 0 \text{ ms}^{-2} $$
(b)(i)
$$ \begin{aligned} a &= 6 \cos 2t \\ v &= \int 6 \cos 2t \text{ d}t = 3 \sin 2t + c \end{aligned} $$
When $t = 0$, $v = 10$:
$$ \begin{aligned} 10 &= 3(0) + c \implies c = 10 \\ \therefore\quad v &= 3 \sin 2t + 10 \end{aligned} $$
(b)(ii)
$$ \begin{aligned} s &= \int (3 \sin 2t + 10) \text{ d}t \\ s &= -\frac{3}{2} \cos 2t + 10t + d \end{aligned} $$
When $t = 0$, $s = 0$:
$$ \begin{aligned} 0 &= -1.5(1) + 0 + d \implies d = 1.5 \\ \therefore\quad s &= -1.5 \cos 2t + 10t + 1.5 \end{aligned} $$
25.
[Source: 0606/13/O/N/20 - Question No. 4]
- The diagram shows the $x-t$ graph for a runner, where displacement, $x$, is measured in metres and time, $t$, is measured in seconds.
- On the axes below, draw the $v-t$ graph for the runner.
- Find the total distance covered by the runner in $125$ s.
- On the axes below, draw the $v-t$ graph for the runner.
- The displacement, $x$ m, of a particle from a fixed point at time $t$ s is given by $x = 6 \cos\left(3t+\frac{\pi}{3}\right)$. Find the acceleration of the particle when $t = \frac{2\pi}{3}$.
(a)(i)
Velocity is the gradient of the $x-t$ graph.
$$ \begin{aligned} 0 \le t \le 50&: \quad v = \frac{150 - 0}{50} = 3 \text{ ms}^{-1} \\ 50 \le t \le 65&: \quad v = 0 \text{ ms}^{-1} \\ 65 \le t \le 85&: \quad v = \frac{50 - 150}{85 - 65} = \frac{-100}{20} = -5 \text{ ms}^{-1} \\ 85 \le t \le 125&: \quad v = 0 \text{ ms}^{-1} \end{aligned} $$
(a)(ii)
Total distance is the sum of magnitudes of changes in displacement.
$$ \begin{aligned} \text{Distance} &= |150 - 0| + |50 - 150| \\ &= 150 + 100 = 250 \text{ m} \end{aligned} $$
(b)
$$ \begin{aligned} x &= 6 \cos\left(3t+\frac{\pi}{3}\right) \\ v &= \frac{\text{d}x}{\text{d}t} = -18 \sin\left(3t+\frac{\pi}{3}\right) \\ a &= \frac{\text{d}v}{\text{d}t} = -54 \cos\left(3t+\frac{\pi}{3}\right) \end{aligned} $$
When $t = \frac{2\pi}{3}$:
$$ \begin{aligned} a &= -54 \cos\left(3\left(\frac{2\pi}{3}\right) + \frac{\pi}{3}\right) \\ &= -54 \cos\left(2\pi + \frac{\pi}{3}\right) \\ &= -54 \cos\left(\frac{\pi}{3}\right) = -54(0.5) = -27 \text{ ms}^{-2} \end{aligned} $$
26.
[Source: 0606/21/O/N/20 - Question No. 9]
A particle moves in a straight line such that, $t$ seconds after passing a fixed point $O$, its displacement from $O$ is $s$ m, where $s = e^{2t} - 10e^{t} - 12t + 9$.
- Find expressions for the velocity and acceleration at time $t$.
- Find the time when the particle is instantaneously at rest.
- Find the acceleration at this time.
(a)
$$ \begin{aligned} v &= \frac{\text{d}s}{\text{d}t} = 2e^{2t} - 10e^{t} - 12 \\ a &= \frac{\text{d}v}{\text{d}t} = 4e^{2t} - 10e^{t} \end{aligned} $$
(b)
For instantaneous rest, $v = 0$:
$$ \begin{aligned} 2e^{2t} - 10e^{t} - 12 &= 0 \\ e^{2t} - 5e^{t} - 6 &= 0 \\ (e^{t} - 6)(e^{t} + 1) &= 0 \end{aligned} $$
Since $e^{t} > 0$, $e^{t} = -1$ has no real solution.
$$ \begin{aligned} e^{t} &= 6 \\ \therefore\quad t &= \ln 6 \text{ s} \end{aligned} $$
(c)
Substitute $e^t = 6$ (and $e^{2t} = 36$) into the acceleration equation:
$$ \begin{aligned} a &= 4(36) - 10(6) \\ &= 144 - 60 = 84 \text{ ms}^{-2} \end{aligned} $$
27.
[Source: 0606/22/F/M/21 - Question No. 12]
A particle $P$ travels in a straight line so that, $t$ seconds after passing through a fixed point $O$, its velocity, $v \text{ ms}^{-1}$ is given by
$$
\begin{array}{ll}
v = \frac{t}{2e} & \text{for } 0 \le t \le 2, \\[0.3cm]
v = e^{-\frac{t}{2}} & \text{for } t > 2.
\end{array}
$$
Given that, after leaving $O$, particle $P$ is never at rest, find the distance it travels between $t = 1$ and $t = 3$.
Since the particle is never at rest (velocity is always positive), distance is simply the integral of velocity over the given time interval.
$$ \text{Distance} = \int_{1}^{3} v \text{ d}t = \int_{1}^{2} \frac{t}{2e} \text{ d}t + \int_{2}^{3} e^{-t/2} \text{ d}t $$
First integral:
$$ \begin{aligned} \int_{1}^{2} \frac{t}{2e} \text{ d}t &= \left[ \frac{t^2}{4e} \right]_{1}^{2} \\ &= \frac{4}{4e} - \frac{1}{4e} = \frac{3}{4e} \end{aligned} $$
Second integral:
$$ \begin{aligned} \int_{2}^{3} e^{-t/2} \text{ d}t &= \left[ -2e^{-t/2} \right]_{2}^{3} \\ &= \left(-2e^{-1.5}\right) - \left(-2e^{-1}\right) \\ &= \frac{2}{e} - 2e^{-1.5} \end{aligned} $$
Total Distance:
$$ \begin{aligned} \text{Distance} &= \frac{3}{4e} + \frac{2}{e} - 2e^{-1.5} \\ &= \frac{11}{4e} - 2e^{-1.5} \text{ m} \end{aligned} $$
28.
[Source: 0606/11/M/J/21 - Question No. 8]
- In this question, all lengths are in metres and time, $t$, is in seconds. The diagram shows the displacement-time graph for a runner, for $0\le t\le40$.
- Find the distance the runner has travelled when $t=40$.
- On the axes below, draw the corresponding velocity-time graph for the runner, for $0\le t\le40$.
- A particle, $P$, moves in a straight line such that its displacement from a fixed point at time $t$ is $s$. The acceleration of $P$ is given by $(2t+4)^{-\frac{1}{2}}$, for $t>0$.
- Given that $P$ has a velocity of $9$ when $t=6$, find the velocity of $P$ at time $t$.
- Given that $s=\frac{1}{3}$ when $t=6$, find the displacement of $P$ at time $t$.
(a)(i)
Total distance = sum of path lengths:
$$ \begin{aligned} \text{Distance} &= |50 - 0| + |-10 - 50| \\ &= 50 + 60 = 110 \text{ m} \end{aligned} $$
(a)(ii)
Velocity is the gradient of the displacement graph.
$$ \begin{aligned} 0 \le t \le 10&: \quad v = \frac{50 - 0}{10 - 0} = 5 \text{ ms}^{-1} \\ 10 < t \le 40&: \quad v = \frac{-10 - 50}{40 - 10} = \frac{-60}{30} = -2 \text{ ms}^{-1} \end{aligned} $$
(b)(i)
$$ \begin{aligned} a &= (2t+4)^{-1/2} \\ v &= \int (2t+4)^{-1/2} \text{ d}t = \frac{(2t+4)^{1/2}}{1/2 \times 2} + c = (2t+4)^{1/2} + c \end{aligned} $$
When $t = 6$, $v = 9$:
$$ \begin{aligned} 9 &= (2(6)+4)^{1/2} + c \\ 9 &= \sqrt{16} + c \implies 9 = 4 + c \implies c = 5 \\ \therefore\quad v &= (2t+4)^{1/2} + 5 \end{aligned} $$
(b)(ii)
$$ \begin{aligned} s &= \int \left( (2t+4)^{1/2} + 5 \right) \text{ d}t \\ &= \frac{(2t+4)^{3/2}}{3/2 \times 2} + 5t + d = \frac{1}{3}(2t+4)^{3/2} + 5t + d \end{aligned} $$
When $t = 6$, $s = \frac{1}{3}$:
$$ \begin{aligned} \frac{1}{3} &= \frac{1}{3}(16)^{3/2} + 5(6) + d \\ \frac{1}{3} &= \frac{1}{3}(64) + 30 + d \\ \frac{1}{3} &= \frac{64}{3} + \frac{90}{3} + d \implies d = \frac{1}{3} - \frac{154}{3} = -51 \\ \therefore\quad s &= \frac{1}{3}(2t+4)^{3/2} + 5t - 51 \end{aligned} $$
29.
[Source: 0606/23/M/J/21 - Question No. 12]
A particle moves in a straight line such that its displacement, $s$ metres, from a fixed point $O$ at time $t$ seconds, is given by $s = 2 + t - 2 \cos t$, for $t \ge 0$.
- Find the displacement of the particle from $O$ at the time when it first comes to instantaneous rest.
- Find the time when the particle next comes to rest.
- Find the distance travelled by the particle for $0 \le t \le \frac{3\pi}{2}$.
(a)
$$ \begin{aligned} v &= \frac{\text{d}s}{\text{d}t} = 1 + 2 \sin t \end{aligned} $$
For instantaneous rest, $v = 0$:
$$ \begin{aligned} 1 + 2 \sin t &= 0 \implies \sin t = -0.5 \end{aligned} $$
First positive value is in the 3rd quadrant:
$$ \begin{aligned} t &= \pi + \frac{\pi}{6} = \frac{7\pi}{6} \text{ s} \end{aligned} $$
Displacement at $t = \frac{7\pi}{6}$:
$$ \begin{aligned} s &= 2 + \frac{7\pi}{6} - 2 \cos\left(\frac{7\pi}{6}\right) \\ &= 2 + \frac{7\pi}{6} - 2\left(-\frac{\sqrt{3}}{2}\right) \\ &= 2 + \frac{7\pi}{6} + \sqrt{3} \text{ m} \end{aligned} $$
(b)
Next time at rest is in the 4th quadrant:
$$ \begin{aligned} t &= 2\pi - \frac{\pi}{6} = \frac{11\pi}{6} \text{ s} \end{aligned} $$
(c)
Between $t = 0$ and $t = \frac{3\pi}{2}$, the particle changes direction at $t = \frac{7\pi}{6}$.
$$ \begin{aligned} s(0) &= 2 + 0 - 2 \cos(0) = 2 - 2 = 0 \\ s\left(\frac{7\pi}{6}\right) &= 2 + \frac{7\pi}{6} + \sqrt{3} \\ s\left(\frac{3\pi}{2}\right) &= 2 + \frac{3\pi}{2} - 2 \cos\left(\frac{3\pi}{2}\right) = 2 + \frac{3\pi}{2} \end{aligned} $$
$$ \begin{aligned} \text{Distance} &= \left| s\left(\frac{7\pi}{6}\right) - s(0) \right| + \left| s\left(\frac{3\pi}{2}\right) - s\left(\frac{7\pi}{6}\right) \right| \\ &= \left(2 + \frac{7\pi}{6} + \sqrt{3} - 0\right) + \left(2 + \frac{7\pi}{6} + \sqrt{3} - \left(2 + \frac{3\pi}{2}\right)\right) \\ &= 2 + \frac{7\pi}{6} + \sqrt{3} + \frac{7\pi}{6} + \sqrt{3} - \frac{3\pi}{2} \\ &= 2 + 2\sqrt{3} + \frac{14\pi}{6} - \frac{9\pi}{6} \\ &= 2 + 2\sqrt{3} + \frac{5\pi}{6} \text{ m} \end{aligned} $$
30.
[Source: 0606/12/O/N/21 - Question No. 11]
- The diagram shows the velocity-time graph for a particle $P$, travelling in a straight line with velocity $v \text{ ms}^{-1}$ at a time $t$ seconds. $P$ accelerates at a constant rate for the first $10$ s of its motion, and then travels at constant velocity, $30 \text{ ms}^{-1}$, for another $15$ s. $P$ then accelerates at a constant rate for a further $10$ s and reaches a velocity of $60 \text{ ms}^{-1}$. $P$ then decelerates at a constant rate and comes to rest when $t=55$.
- Find the acceleration when $t=12$.
- Find the acceleration when $t=50$.
- Find the total distance travelled by the particle $P$.
- A particle $Q$ travels in a straight line such that its velocity, $v \text{ ms}^{-1}$, at time $t$ s after passing through a fixed point $O$ is given by $v = 4 \cos 3t - 4$.
- Find the speed of $Q$ when $t = \frac{5\pi}{9}$.
- Find the smallest positive value of $t$ for which the acceleration of $Q$ is zero.
- Find an expression for the displacement of $Q$ from $O$ at time $t$.
(a)(i)
At $t=12$, the particle is travelling at constant velocity ($30 \text{ ms}^{-1}$).
$$ \therefore\quad \text{Acceleration } = 0 \text{ ms}^{-2} $$
(a)(ii)
At $t=50$, it is on the line segment from $t=35, v=60$ to $t=55, v=0$.
$$ \begin{aligned} a &= \text{gradient} = \frac{0 - 60}{55 - 35} = \frac{-60}{20} = -3 \text{ ms}^{-2} \end{aligned} $$
(a)(iii)
Total distance is the total area under the $v-t$ graph.
$$ \begin{aligned} \text{Area}_{0-10} &= \frac{1}{2} \times 10 \times 30 = 150 \\ \text{Area}_{10-25} &= 15 \times 30 = 450 \\ \text{Area}_{25-35} &= \frac{1}{2}(30 + 60) \times 10 = 450 \\ \text{Area}_{35-55} &= \frac{1}{2} \times 20 \times 60 = 600 \\ \text{Total Distance} &= 150 + 450 + 450 + 600 = 1650 \text{ m} \end{aligned} $$
(b)(i)
$$ \begin{aligned} v &= 4 \cos\left(3 \times \frac{5\pi}{9}\right) - 4 \\ &= 4 \cos\left(\frac{5\pi}{3}\right) - 4 \\ &= 4(0.5) - 4 = 2 - 4 = -2 \text{ ms}^{-1} \end{aligned} $$
Speed is the magnitude of velocity.
$$ \therefore\quad \text{Speed} = |-2| = 2 \text{ ms}^{-1} $$
(b)(ii)
$$ \begin{aligned} a &= \frac{\text{d}v}{\text{d}t} = -12 \sin 3t \end{aligned} $$
For zero acceleration:
$$ \begin{aligned} -12 \sin 3t &= 0 \implies \sin 3t = 0 \end{aligned} $$
Smallest positive value:
$$ \begin{aligned} 3t &= \pi \implies t = \frac{\pi}{3} \text{ s} \end{aligned} $$
(b)(iii)
$$ \begin{aligned} s &= \int (4 \cos 3t - 4) \text{ d}t = \frac{4}{3} \sin 3t - 4t + c \end{aligned} $$
When $t = 0$, $s = 0$:
$$ \begin{aligned} 0 &= 0 - 0 + c \implies c = 0 \\ \therefore\quad s &= \frac{4}{3} \sin 3t - 4t \end{aligned} $$
31.
[Source: 0606/22/F/M/22 - Question No. 10]
- A vehicle travels along a straight, horizontal road. At time $t=0$ seconds, the vehicle, travelling at a velocity of $w \text{ ms}^{-1}$ passes point $O$. The vehicle travels at this constant velocity for $12$ seconds. It then slows down, with constant deceleration, for $10$ seconds until it reaches a velocity of $(w-14) \text{ ms}^{-1}$. It continues to travel at this velocity for $28$ seconds until it reaches point $A$, $458$ m from $O$. Find the value of $w$.
- A particle moves in a straight line. The velocity, $v \text{ ms}^{-1}$, of the particle at time $t$ seconds, where $t\ge0$, is given by $v=(t-4)(t-5)$.
- Find the value of $t$ for which the acceleration of the particle is $0 \text{ ms}^{-2}$.
- Find the set of values of $t$ for which the velocity of the particle is negative.
- Find the distance travelled by the particle in the first $5$ seconds of its motion.
(a)
Total distance is the area under the velocity-time graph. The motion has three parts: a rectangle, a trapezium, and a rectangle.
$$ \begin{aligned} \text{Area}_1 &= 12 \times w = 12w \\ \text{Area}_2 &= \frac{1}{2} \times 10 \times (w + w - 14) = 5(2w - 14) = 10w - 70 \\ \text{Area}_3 &= 28 \times (w - 14) = 28w - 392 \\ \text{Total Distance} &= 12w + 10w - 70 + 28w - 392 = 458 \\ 50w - 462 &= 458 \\ 50w &= 920 \\ \therefore\quad w &= 18.4 \end{aligned} $$
(b)(i)
$$ \begin{aligned} v &= (t-4)(t-5) = t^2 - 9t + 20 \\ a &= \frac{\text{d}v}{\text{d}t} = 2t - 9 \end{aligned} $$
When $a = 0$:
$$ 2t - 9 = 0 \implies t = 4.5 \text{ s} $$
(b)(ii)
For negative velocity:
$$ \begin{aligned} (t-4)(t-5) &< 0 \\ \therefore\quad 4 &< t < 5 \end{aligned} $$
(b)(iii)
Displacement $s = \int (t^2 - 9t + 20) \text{ d}t = \frac{t^3}{3} - \frac{9t^2}{2} + 20t + c$. Let $s(0) = 0$.
The particle changes direction at $t=4$ and $t=5$.
$$ \begin{aligned} s(4) &= \frac{64}{3} - \frac{9(16)}{2} + 80 = 21.333 - 72 + 80 = 29.333 = \frac{88}{3} \\ s(5) &= \frac{125}{3} - \frac{9(25)}{2} + 100 = 41.667 - 112.5 + 100 = 29.167 = \frac{175}{6} \end{aligned} $$
Distance = Distance from $t=0$ to $4$ + Distance from $t=4$ to $5$:
$$ \begin{aligned} \text{Distance} &= |s(4) - s(0)| + |s(5) - s(4)| \\ &= \frac{88}{3} + \left| \frac{175}{6} - \frac{176}{6} \right| = \frac{176}{6} + \frac{1}{6} = \frac{177}{6} = 29.5 \text{ m} \end{aligned} $$
32.
[Source: 0606/11/M/J/22 - Question No. 2]
A particle moves in a straight line such that its displacement, $s$ metres, from a fixed point, at time $t$ seconds, $t\ge0$, is given by $s=(1+3t)^{-\frac{1}{2}}$.
- Find the exact speed of the particle when $t=1$.
- Show that the acceleration of the particle will never be zero.
(a)
$$ \begin{aligned} v &= \frac{\text{d}s}{\text{d}t} = -\frac{1}{2}(1+3t)^{-\frac{3}{2}} \times 3 = -\frac{3}{2}(1+3t)^{-\frac{3}{2}} \end{aligned} $$
When $t=1$:
$$ \begin{aligned} v &= -\frac{3}{2}(1+3)^{-\frac{3}{2}} = -\frac{3}{2}(4)^{-\frac{3}{2}} = -\frac{3}{2}\left(\frac{1}{8}\right) = -\frac{3}{16} \text{ ms}^{-1} \end{aligned} $$
Speed is magnitude of velocity:
$$ \therefore\quad \text{Speed} = \frac{3}{16} \text{ ms}^{-1} $$
(b)
$$ \begin{aligned} a &= \frac{\text{d}v}{\text{d}t} = -\frac{3}{2} \times \left(-\frac{3}{2}\right) (1+3t)^{-\frac{5}{2}} \times 3 = \frac{27}{4}(1+3t)^{-\frac{5}{2}} \\ a &= \frac{27}{4(1+3t)^{\frac{5}{2}}} \end{aligned} $$
For all $t \ge 0$, $(1+3t) \ge 1 > 0$, so $(1+3t)^{\frac{5}{2}} > 0$.
Since the numerator is $27 \neq 0$ and the denominator is strictly positive, the fraction is always positive.
Therefore, the acceleration will never be zero.
33.
[Source: 0606/12/M/J/22 - Question No. 11]
In this question all lengths are in kilometres and time is in hours. A particle P moves in a straight line such that its displacement, $s$, from a fixed point at time $t$ is given by $s=(t+2)(t-5)^{2}$ for $t\ge0$.
- Find the values of $t$ for which the velocity of P is zero.
- On the axes, draw the displacement-time graph for P for $0\le t\le6$, stating the coordinates of the points where the graph meets the coordinate axes.
- On the axes below, draw the velocity-time graph for P for $0\le t\le6$, stating the coordinates of the points where the graph meets the coordinate axes.
-
- Write down an expression for the acceleration of P at time $t$.
- Hence, on the axes below, draw the acceleration-time graph for P for $0\le t\le6$, stating the coordinates of the points where the graph meets the coordinate axes.
(a)
$$ \begin{aligned} s &= (t+2)(t-5)^2 \\ v &= \frac{\text{d}s}{\text{d}t} = 1 \cdot (t-5)^2 + (t+2) \cdot 2(t-5) \\ &= (t-5) [ (t-5) + 2(t+2) ] \\ &= (t-5)(3t - 1) \end{aligned} $$
When $v = 0$:
$$ \begin{aligned} t - 5 = 0 \quad &\text{or} \quad 3t - 1 = 0 \\ t = 5 \text{ hours} \quad &\text{or} \quad t = \frac{1}{3} \text{ hours} \end{aligned} $$
(d)(i)
$$ \begin{aligned} v &= (t-5)(3t-1) = 3t^2 - 16t + 5 \\ a &= \frac{\text{d}v}{\text{d}t} = 6t - 16 \end{aligned} $$
34.
[Source: 0606/13/M/J/22 - Question No. 7]
The velocity, $v \text{ ms}^{-1}$, of a particle moving in a straight line, $t$ seconds after passing through a fixed point O, is given by $v=6 \sin 3t$.
- Find the time at which the acceleration of the particle is first equal to $-9 \text{ ms}^{-2}$.
- Find the displacement of the particle from O when $t=5.6$.
(a)
$$ \begin{aligned} a &= \frac{\text{d}v}{\text{d}t} = 18 \cos 3t \end{aligned} $$
When $a = -9$:
$$ \begin{aligned} 18 \cos 3t &= -9 \\ \cos 3t &= -0.5 \end{aligned} $$
First positive value:
$$ \begin{aligned} 3t &= \frac{2\pi}{3} \\ \therefore\quad t &= \frac{2\pi}{9} \text{ s} \quad (\approx 0.698 \text{ s}) \end{aligned} $$
(b)
$$ \begin{aligned} s &= \int 6 \sin 3t \text{ d}t = -2 \cos 3t + c \end{aligned} $$
Passing through $O$ means $s = 0$ when $t = 0$:
$$ \begin{aligned} 0 &= -2(1) + c \implies c = 2 \\ \therefore\quad s &= 2 - 2 \cos 3t \end{aligned} $$
When $t = 5.6$:
$$ \begin{aligned} s &= 2 - 2 \cos(3 \times 5.6) = 2 - 2 \cos(16.8) \\ &\approx 2 - 2(-0.4755) \approx 2 + 0.951 \\ &= 2.95 \text{ m} \end{aligned} $$
35.
[Source: 0606/12/O/N/22 - Question No. 11]
A particle $P$ moves in a straight line such that, $t$ seconds after passing through a fixed point $O$, its displacement, $s$ metres, is given by
$$s=\frac{(2t+1)^{\frac{3}{2}}}{t+1}-1.$$
- Show that the velocity of $P$ at time $t$ can be written in the form $\frac{(2t+1)^{\frac{1}{2}}}{(t+1)^{2}}(a+bt)$, where $a$ and $b$ are integers to be found.
- Show that $P$ is never at instantaneous rest after passing through $O$.
(a)
Using Quotient Rule for $s = u/v$ where $u = (2t+1)^{3/2}, v = t+1$:
$$ \begin{aligned} \frac{\text{d}u}{\text{d}t} &= \frac{3}{2}(2t+1)^{1/2}(2) = 3(2t+1)^{1/2} \\ \frac{\text{d}v}{\text{d}t} &= 1 \\ v &= \frac{v u' - u v'}{v^2} = \frac{(t+1)\left[3(2t+1)^{1/2}\right] - (2t+1)^{3/2}(1)}{(t+1)^2} \\ &= \frac{(2t+1)^{1/2} [ 3(t+1) - (2t+1) ]}{(t+1)^2} \\ &= \frac{(2t+1)^{1/2} (3t + 3 - 2t - 1)}{(t+1)^2} \\ &= \frac{(2t+1)^{1/2}}{(t+1)^2} (2 + t) \end{aligned} $$
Comparing this with $\frac{(2t+1)^{1/2}}{(t+1)^2}(a+bt)$, we get $a=2, b=1$.
(b)
For $t \ge 0$:
$(2t+1)^{1/2} \ge 1 > 0$ and $(t+1)^2 \ge 1 > 0$.
$(2+t) \ge 2 > 0$.
Therefore, the velocity $v > 0$ for all $t \ge 0$. The particle is never at rest.
36.
[Source: 0606/21/O/N/22 - Question No. 10]
The acceleration, $a \text{ ms}^{-2}$, of a particle at time $t$ seconds is given by $a=-\frac{45}{(t+1)^{2}}$. When $t=0$ the velocity of the particle is $50 \text{ ms}^{-1}$.
- Find an expression for the velocity of the particle in terms of $t$.
- Find the distance travelled by the particle between $t=1$ and $t=10$.
(a)
$$ \begin{aligned} v &= \int -45(t+1)^{-2} \text{ d}t = \frac{-45(t+1)^{-1}}{-1} + c = \frac{45}{t+1} + c \end{aligned} $$
When $t = 0$, $v = 50$:
$$ \begin{aligned} 50 &= 45 + c \implies c = 5 \\ \therefore\quad v &= \frac{45}{t+1} + 5 \end{aligned} $$
(b)
Since $v > 0$ for all $t \ge 0$, the distance is the definite integral of velocity.
$$ \begin{aligned} s &= \int \left( \frac{45}{t+1} + 5 \right) \text{ d}t = 45 \ln(t+1) + 5t \end{aligned} $$
Evaluate between $t=1$ and $t=10$:
$$ \begin{aligned} \text{Distance} &= [45 \ln 11 + 50] - [45 \ln 2 + 5] \\ &= 45 \ln 11 - 45 \ln 2 + 45 \\ &= 45 \ln\left(\frac{11}{2}\right) + 45 \\ &= 45 \ln 5.5 + 45 \approx 121.7 \text{ m} \end{aligned} $$
37.
[Source: 0606/23/O/N/22 - Question No. 8]
- Particle $A$ starts from the point with position vector $\binom{3}{-2}$ and travels with speed $26 \text{ ms}^{-1}$ in the direction of the vector $\binom{12}{5}$. Find the position vector of $A$ after $t$ seconds.
- At the same time, particle $B$ starts from the point with position vector $\binom{67}{-18}$. It travels with speed $20 \text{ ms}^{-1}$ at an angle of $\alpha$ above the positive $x$-axis, where $\tan \alpha = \frac{3}{4}$. Find the position vector of $B$ after $t$ seconds.
- Hence find the time at which $A$ and $B$ meet, and the position where this occurs.
(a)
Direction magnitude = $\sqrt{12^2 + 5^2} = 13$.
Velocity vector of A = $\frac{26}{13} \binom{12}{5} = 2 \binom{12}{5} = \binom{24}{10}$.
$$ r_A = \binom{3}{ -2} + t \binom{24}{10} = \binom{3 + 24t}{-2 + 10t} $$
Velocity vector of A = $\frac{26}{13} \binom{12}{5} = 2 \binom{12}{5} = \binom{24}{10}$.
$$ r_A = \binom{3}{ -2} + t \binom{24}{10} = \binom{3 + 24t}{-2 + 10t} $$
(b)
$\tan \alpha = \frac{3}{4} \implies \sin \alpha = \frac{3}{5}, \cos \alpha = \frac{4}{5}$.
Velocity vector of B = $\binom{20 \cos \alpha}{20 \sin \alpha} = \binom{20(4/5)}{20(3/5)} = \binom{16}{12}$.
$$ r_B = \binom{67}{-18} + t \binom{16}{12} = \binom{67 + 16t}{-18 + 12t} $$
Velocity vector of B = $\binom{20 \cos \alpha}{20 \sin \alpha} = \binom{20(4/5)}{20(3/5)} = \binom{16}{12}$.
$$ r_B = \binom{67}{-18} + t \binom{16}{12} = \binom{67 + 16t}{-18 + 12t} $$
(c)
Equate $x$-components:
$$ \begin{aligned} 3 + 24t &= 67 + 16t \\ 8t &= 64 \implies t = 8 \text{ s} \end{aligned} $$
Check with $y$-components:
$$ -2 + 10(8) = 78 \quad \text{and} \quad -18 + 12(8) = 78 \quad (\text{They match}) $$
$$ \therefore\quad \text{Position} = \binom{195}{78} $$
38.
[Source: 0606/12/F/M/23 - Question No. 9]
In this question, all lengths are in metres. A particle $P$ has position vector $\binom{2+12t}{5-5t}$ at a time $t$ seconds, $t\ge0$.
- Write down the initial position vector of $P$.
- Find the speed of $P$.
- Determine whether $P$ passes through the point with position vector $\binom{158}{-48}$.
(a)
When $t = 0$:
$$ \text{Initial Position} = \binom{2}{5} $$
(b)
Velocity vector is the coefficient of $t$, which is $\binom{12}{-5}$.
$$ \text{Speed} = \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = 13 \text{ ms}^{-1} $$
(c)
Equate $x$-component:
$$ \begin{aligned} 2 + 12t &= 158 \\ 12t &= 156 \implies t = 13 \end{aligned} $$
Check $y$-component at $t=13$:
$$ 5 - 5(13) = 5 - 65 = -60 \neq -48 $$
Since the components do not match at the same time $t$, $P$ does NOT pass through $\binom{158}{-48}$.
39.
[Source: 0606/22/F/M/23 - Question No. 10]
A particle $P$ moves in a straight line such that, $t$ seconds after passing a fixed point $O$, its acceleration, $a \text{ ms}^{-2}$ is given by
$$ \begin{aligned} a &= 6t \quad \text{for } 0 \le t \le 3 \\ a &= \frac{18e^3}{e^t} \quad \text{for } t \ge 3 \end{aligned} $$
When $t=1$, the velocity of $P$ is $2 \text{ ms}^{-1}$ and its displacement from $O$ is $-4\text{ m}$.
-
- Find the velocity of $P$ when $t=3$.
- Find the displacement of $P$ from $O$ when $t=3$.
- Find an expression in terms of $t$ for the displacement of $P$ from $O$ when $t \ge 3$.
(a)(i)
For $0 \le t \le 3$:
$$ \begin{aligned} v &= \int 6t \text{ d}t = 3t^2 + c \end{aligned} $$
When $t=1$, $v=2$:
$$ \begin{aligned} 2 &= 3(1)^2 + c \implies c = -1 \\ \therefore\quad v &= 3t^2 - 1 \end{aligned} $$
When $t=3$:
$$ v(3) = 3(9) - 1 = 26 \text{ ms}^{-1} $$
(a)(ii)
$$ \begin{aligned} s &= \int (3t^2 - 1) \text{ d}t = t^3 - t + d \end{aligned} $$
When $t=1$, $s=-4$:
$$ \begin{aligned} -4 &= 1^3 - 1 + d \implies d = -4 \\ \therefore\quad s &= t^3 - t - 4 \end{aligned} $$
When $t=3$:
$$ s(3) = 3^3 - 3 - 4 = 27 - 7 = 20 \text{ m} $$
(b)
For $t \ge 3$, $a = 18e^{3-t}$:
$$ \begin{aligned} v &= \int 18e^{3-t} \text{ d}t = -18e^{3-t} + C \end{aligned} $$
By continuity, $v(3) = 26$:
$$ \begin{aligned} 26 &= -18e^0 + C \implies C = 44 \\ \therefore\quad v &= 44 - 18e^{3-t} \end{aligned} $$
$$ \begin{aligned} s &= \int (44 - 18e^{3-t}) \text{ d}t = 44t + 18e^{3-t} + D \end{aligned} $$
By continuity, $s(3) = 20$:
$$ \begin{aligned} 20 &= 44(3) + 18e^0 + D \\ 20 &= 132 + 18 + D \implies D = -130 \\ \therefore\quad s &= 44t + 18e^{3-t} - 130 \end{aligned} $$
40.
[Source: 0606/11/M/J/23 - Question No. 7]
The diagram shows the velocity-time graph for a particle travelling in a straight line with velocity, $v \text{ ms}^{-1}$, at time $t$ seconds. When $t=30$ the velocity of the particle is $V \text{ ms}^{-1}$. The particle travels $800$ metres in $45$ seconds.
- Find the value of $V$.
- Find the acceleration of the particle when $t=35$.
(a)
Area under graph = Total Distance = $800$.
$$ \begin{aligned} \text{Area}_1 (\text{rectangle}) &= 10 \times 10 = 100 \\ \text{Area}_2 (\text{trapezium}) &= \frac{1}{2}(10+V) \times 20 = 10(10+V) = 100 + 10V \\ \text{Area}_3 (\text{triangle}) &= \frac{1}{2} \times 15 \times V = 7.5V \\ \text{Total} &= 100 + 100 + 10V + 7.5V = 800 \\ 200 + 17.5V &= 800 \\ 17.5V &= 600 \\ \therefore\quad V &= \frac{600}{17.5} = \frac{240}{7} \text{ ms}^{-1} \quad (\approx 34.29 \text{ ms}^{-1}) \end{aligned} $$
(b)
At $t = 35$, acceleration is the gradient of the line from $(30, V)$ to $(45, 0)$.
$$ \begin{aligned} a &= \frac{0 - V}{45 - 30} = \frac{-240/7}{15} \\ &= -\frac{16}{7} \text{ ms}^{-2} \quad (\approx -2.29 \text{ ms}^{-2}) \end{aligned} $$
41.
[Source: 0606/13/M/J/23 - Question No. 14]
In this question lengths are in centimetres and time is in seconds. A particle $P$ moves in a straight line such that its displacement $s$, from a fixed point at a time $t$, is given by $s=3(t+2)(t-4)^2$ for $0 \le t \le 5$.
- Find the values of $t$ for which the velocity, $v$, of $P$ is zero.
- On the axes below, sketch the displacement-time graph of $P$, stating the intercepts with the axes.
- On the axes below, sketch the velocity-time graph of $P$, stating the intercepts with the axes.
-
- Find an expression for the acceleration of $P$ at time $t$.
- Hence, on the axes below, sketch the acceleration-time graph of $P$, stating the intercepts with the axes.
(a)
$$ \begin{aligned} s &= 3(t+2)(t-4)^2 \\ v &= \frac{\text{d}s}{\text{d}t} = 3 \left[ 1 \cdot (t-4)^2 + (t+2) \cdot 2(t-4) \right] \\ &= 3(t-4) [ (t-4) + 2(t+2) ] \\ &= 3(t-4) [ 3t ] = 9t(t-4) \end{aligned} $$
For $v = 0$:
$$ 9t(t-4) = 0 \implies t = 0 \text{ s} \quad \text{or} \quad t = 4 \text{ s} $$
(b)
Intercepts for $s-t$ graph: At $t=0, s=3(2)(16)=96$. At $t=4, s=0$. At $t=5, s=3(7)(1)=21$.
(c)
Intercepts for $v-t$ graph: $v = 9t^2 - 36t$. At $t=0, v=0$. At $t=4, v=0$. Min at $t=2, v=-36$. At $t=5, v=45$.
(d)(i)
$$ \begin{aligned} v &= 9t^2 - 36t \\ a &= \frac{\text{d}v}{\text{d}t} = 18t - 36 \end{aligned} $$
(d)(ii)
Intercepts for $a-t$ graph: $a = 18t - 36$. At $t=0, a=-36$. At $t=2, a=0$. At $t=5, a=54$.
42.
[Source: 0606/22/M/J/23 - Question No. 13]
A particle travels in a straight line so that, $t$ seconds after passing a fixed point, its velocity, $v \text{ ms}^{-1}$ is given by
$$ \begin{aligned} v &= e^{\frac{t}{4}} \quad \text{for } 0 \le t \le 4 \\ v &= \frac{16e}{t^2} \quad \text{for } 4 \le t \le k \end{aligned} $$
The total distance travelled by the particle between $t=0$ and $t=k$ is $13.4$ metres. Find the value of $k$.
Since $v > 0$ for all $t \ge 0$, distance is the integral of velocity.
Distance for first part ($0 \le t \le 4$):
$$ \begin{aligned} D_1 &= \int_{0}^{4} e^{t/4} \text{ d}t = \left[ \frac{e^{t/4}}{1/4} \right]_{0}^{4} \\ &= \left[ 4e^{t/4} \right]_{0}^{4} = 4e^1 - 4e^0 = 4e - 4 \end{aligned} $$
Distance for second part ($4 \le t \le k$):
$$ \begin{aligned} D_2 &= \int_{4}^{k} 16e \cdot t^{-2} \text{ d}t = \left[ \frac{-16e}{t} \right]_{4}^{k} \\ &= \left( -\frac{16e}{k} \right) - \left( -\frac{16e}{4} \right) = 4e - \frac{16e}{k} \end{aligned} $$
Total Distance = $D_1 + D_2$:
$$ \begin{aligned} (4e - 4) + \left( 4e - \frac{16e}{k} \right) &= 13.4 \\ 8e - 4 - \frac{16e}{k} &= 13.4 \\ 8e - 17.4 &= \frac{16e}{k} \\ k &= \frac{16e}{8e - 17.4} \end{aligned} $$
Using $e \approx 2.71828$:
$$ \begin{aligned} k &= \frac{16(2.71828)}{8(2.71828) - 17.4} = \frac{43.4925}{21.746 - 17.4} = \frac{43.4925}{4.346} \approx 10 \end{aligned} $$
$$ \therefore\quad k = 10 $$
43.
[Source: 0606/21/O/N/24 - Question No. 11]
A particle moves in a straight line so that its displacement from a fixed point $O$ at time $t$ seconds is $x$ metres, where $x=t^{3}+t^{2}-t+8$ and $t\ge0$.
- Find the time when the particle changes direction.
- Show that the particle is moving towards $O$ when $t=0$.
- Find the total distance travelled by the particle during the first $2$ seconds of its motion.
(a)
$$ \begin{aligned} v &= \frac{\text{d}x}{\text{d}t} = 3t^2 + 2t - 1 \end{aligned} $$
Particle changes direction when $v = 0$:
$$ \begin{aligned} 3t^2 + 2t - 1 &= 0 \\ (3t - 1)(t + 1) &= 0 \end{aligned} $$
Since $t \ge 0$, $t = \frac{1}{3} \text{ s}$.
(b)
At $t = 0$:
$$ \begin{aligned} x &= (0)^3 + (0)^2 - (0) + 8 = 8 \text{ m} \\ v &= 3(0)^2 + 2(0) - 1 = -1 \text{ ms}^{-1} \end{aligned} $$
Since its position is positive ($x > 0$) but its velocity is negative ($v < 0$), the particle is moving back towards the origin $O$.
(c)
The particle changes direction at $t = \frac{1}{3}$.
$$ \begin{aligned} x(0) &= 8 \\ x\left(\frac{1}{3}\right) &= \left(\frac{1}{3}\right)^3 + \left(\frac{1}{3}\right)^2 - \left(\frac{1}{3}\right) + 8 = \frac{1}{27} + \frac{3}{27} - \frac{9}{27} + 8 = 8 - \frac{5}{27} = \frac{211}{27} \\ x(2) &= (2)^3 + (2)^2 - (2) + 8 = 8 + 4 - 2 + 8 = 18 \end{aligned} $$
Total distance:
$$ \begin{aligned} \text{Distance} &= \left| x\left(\frac{1}{3}\right) - x(0) \right| + \left| x(2) - x\left(\frac{1}{3}\right) \right| \\ &= \left| \frac{211}{27} - \frac{216}{27} \right| + \left| 18 - \frac{211}{27} \right| \\ &= \frac{5}{27} + \left| \frac{486}{27} - \frac{211}{27} \right| = \frac{5}{27} + \frac{275}{27} = \frac{280}{27} \text{ m} \quad (\approx 10.37 \text{ m}) \end{aligned} $$
44.
[Source: 0606/22/O/N/24 - Question No. 9]
In this question time is measured in seconds.
- A particle is moving in a straight line with constant velocity of $6 \text{ ms}^{-1}$. At time $t=0$, it passes a fixed point $A$. At time $t=5$ it suddenly changes direction and moves with a different constant velocity along the same straight line. It passes the point $A$ again at time $t=15$. Sketch the velocity-time graph for the motion.
- Another particle is moving in a straight line with constant acceleration. At time $t=0$ it passes a fixed point $B$ with velocity $-8 \text{ ms}^{-1}$. It passes the point $B$ again at time $t=20$. Sketch the velocity-time graph for the motion.
(a)
For $0 \le t \le 5$, the velocity is $6 \text{ ms}^{-1}$.
Distance travelled = $5 \times 6 = 30 \text{ m}$.
To return to $A$, the particle must travel $-30 \text{ m}$ over the next $10$ seconds ($15 - 5$).
New constant velocity = $\frac{-30}{10} = -3 \text{ ms}^{-1}$.
Distance travelled = $5 \times 6 = 30 \text{ m}$.
To return to $A$, the particle must travel $-30 \text{ m}$ over the next $10$ seconds ($15 - 5$).
New constant velocity = $\frac{-30}{10} = -3 \text{ ms}^{-1}$.
(b)
Constant acceleration means the $v-t$ graph is a straight line.
It passes $B$ at $t=0$ and again at $t=20$. This means total displacement is $0$.
Therefore, the area of the triangle below the $t$-axis must equal the area above the $t$-axis. This happens when the line crosses the axis exactly halfway, at $t = 10$.
At $t = 20$, the velocity must be $+8 \text{ ms}^{-1}$.
It passes $B$ at $t=0$ and again at $t=20$. This means total displacement is $0$.
Therefore, the area of the triangle below the $t$-axis must equal the area above the $t$-axis. This happens when the line crosses the axis exactly halfway, at $t = 10$.
At $t = 20$, the velocity must be $+8 \text{ ms}^{-1}$.
45.
[Source: 0606/23/O/N/24 - Question No. 12]
A particle moves in a straight line. Its velocity, $v \text{ ms}^{-1}$, at time $t$ seconds is given by $v=\cos t - \sin t$.
- Find the acceleration, $a \text{ ms}^{-2}$, when $t=\frac{\pi}{3}$.
- Find the displacement at the time when the particle first changes direction after passing through $O$.
- Find an expression for $a$ in terms of $s$.
(a)
$$ \begin{aligned} a &= \frac{\text{d}v}{\text{d}t} = -\sin t - \cos t \end{aligned} $$
When $t = \frac{\pi}{3}$:
$$ \begin{aligned} a &= -\sin\left(\frac{\pi}{3}\right) - \cos\left(\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2} - \frac{1}{2} = -\frac{1+\sqrt{3}}{2} \text{ ms}^{-2} \end{aligned} $$
(b)
Particle changes direction when $v = 0$:
$$ \begin{aligned} \cos t - \sin t &= 0 \implies \tan t = 1 \end{aligned} $$
First positive value is $t = \frac{\pi}{4}$.
$$ \begin{aligned} s &= \int (\cos t - \sin t) \text{ d}t = \sin t + \cos t + c \end{aligned} $$
At $t = 0, s = 0$:
$$ \begin{aligned} 0 &= \sin(0) + \cos(0) + c \implies 0 = 1 + c \implies c = -1 \\ \therefore\quad s &= \sin t + \cos t - 1 \end{aligned} $$
When $t = \frac{\pi}{4}$:
$$ \begin{aligned} s &= \sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right) - 1 \\ &= \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} - 1 = \sqrt{2} - 1 \text{ m} \end{aligned} $$
(c)
$$ \begin{aligned} a &= -\sin t - \cos t = -(\sin t + \cos t) \end{aligned} $$
From the displacement equation, $s = \sin t + \cos t - 1$, we can rearrange to get:
$$ \sin t + \cos t = s + 1 $$
Substitute this into the expression for $a$:
$$ a = -(s + 1) = -s - 1 $$
46.
[Source: 0606/22/F/M/25 - Question No. 12]
In this question, all lengths are in metres and time is in seconds. A particle, $P$, moves in a straight line such that $t$ seconds after passing through a fixed point $O$ its displacement, $s$, is given by $s=5\ln(2t+1)-5t$.
- Find the value of $t$ for which $P$ is instantaneously at rest.
- Find the distance $P$ travels between $t=0$ and $t=2$.
- Find an expression for the acceleration of $P$ in terms of $t$.
- Find the acceleration when $t=4.5$.
(a)
$$ \begin{aligned} v &= \frac{\text{d}s}{\text{d}t} = 5\left(\frac{2}{2t+1}\right) - 5 = \frac{10}{2t+1} - 5 \end{aligned} $$
For instantaneous rest, $v = 0$:
$$ \begin{aligned} \frac{10}{2t+1} - 5 &= 0 \implies \frac{10}{2t+1} = 5 \implies 2t+1 = 2 \implies t = 0.5 \text{ s} \end{aligned} $$
(b)
The particle changes direction at $t = 0.5$.
$$ \begin{aligned} s(0) &= 5\ln(1) - 0 = 0 \\ s(0.5) &= 5\ln(2(0.5)+1) - 5(0.5) = 5\ln 2 - 2.5 \\ s(2) &= 5\ln(2(2)+1) - 5(2) = 5\ln 5 - 10 \end{aligned} $$
$$ \begin{aligned} \text{Distance} &= |s(0.5) - s(0)| + |s(2) - s(0.5)| \\ &= (5\ln 2 - 2.5) + |5\ln 5 - 10 - (5\ln 2 - 2.5)| \\ &= (5\ln 2 - 2.5) + |5\ln 5 - 5\ln 2 - 7.5| \end{aligned} $$
Since $5\ln\left(\frac{5}{2}\right) - 7.5 < 0$, its absolute value is $7.5 - 5\ln 5 + 5\ln 2$.
$$ \begin{aligned} \text{Distance} &= 5\ln 2 - 2.5 + 7.5 - 5\ln 5 + 5\ln 2 \\ &= 5 + 10\ln 2 - 5\ln 5 \text{ m} = 5 + 5\ln\left(\frac{4}{5}\right) \text{ m} \quad (\approx 3.88 \text{ m}) \end{aligned} $$
(c)
$$ \begin{aligned} v &= 10(2t+1)^{-1} - 5 \\ a &= \frac{\text{d}v}{\text{d}t} = -10(2t+1)^{-2} \times 2 = \frac{-20}{(2t+1)^2} \end{aligned} $$
(d)
When $t = 4.5$:
$$ \begin{aligned} a &= \frac{-20}{(2(4.5)+1)^2} = \frac{-20}{10^2} = \frac{-20}{100} = -0.2 \text{ ms}^{-2} \end{aligned} $$
47.
[Source: 0606/13/M/J/25 - Question No. 11]
A particle $P$ moves in a straight line and passes through a fixed point $O$. At time $t$ seconds, its displacement from $O$, $s$ metres, is given by
$$ \begin{aligned} s &= t+6t^{2}-t^{3} \quad \text{for } 0 \le t \le 3 \\ s &= 12t-\frac{1}{3}t^{2}-3 \quad \text{for } 3 \le t \le k \end{aligned} $$
where $k$ is a constant. It is given that, for $3 \le t \le k$, the velocity of $P$ is positive and its acceleration is negative.
- The maximum velocity of $P$ occurs when $t=2$. On the axes below, sketch a velocity-time graph for the first $k$ seconds of the motion of $P$.
- The total distance travelled by $P$ for $0 \le t \le k$ is $57$ metres. Given that when $t=3$ the distance and displacement of $P$ from $O$ are equal, find the value of $k$.
(a)
For $0 \le t \le 3$, $v = \frac{\text{d}s}{\text{d}t} = 1 + 12t - 3t^2$.
At $t=0, v=1$. At $t=2$ (max), $v = 1 + 24 - 12 = 13$. At $t=3, v = 1 + 36 - 27 = 10$.
For $3 \le t \le k$, $v = 12 - \frac{2}{3}t$.
This is a straight line. At $t=3, v=10$.
At $t=0, v=1$. At $t=2$ (max), $v = 1 + 24 - 12 = 13$. At $t=3, v = 1 + 36 - 27 = 10$.
For $3 \le t \le k$, $v = 12 - \frac{2}{3}t$.
This is a straight line. At $t=3, v=10$.
(b)
Since the distance and displacement are equal at $t=3$, the particle has not reversed direction during $0 \le t \le 3$.
Also, for $3 \le t \le k$, velocity is positive, meaning the particle continues in the same direction.
Therefore, total distance travelled at $t=k$ is equal to its displacement at $t=k$.
$$ \begin{aligned} s(k) &= 57 \\ 12k - \frac{1}{3}k^2 - 3 &= 57 \\ \frac{1}{3}k^2 - 12k + 60 &= 0 \\ k^2 - 36k + 180 &= 0 \\ (k - 6)(k - 30) &= 0 \end{aligned} $$
$k = 6$ or $k = 30$.
However, the velocity must be positive for $3 \le t \le k$:
$$ v = 12 - \frac{2}{3}k > 0 \implies \frac{2}{3}k < 12 \implies k < 18 $$
$$ \therefore\quad k = 6 $$
48.
[Source: 0606/21/M/J/25 - Question No. 10]
A particle $P$ moves in a straight line. $t$ seconds after passing a fixed point, $O$, the acceleration of $P$, $a \text{ ms}^{-2}$ is given by
$$ \begin{aligned} a &= t^{2}-2 \quad \text{for } 0 \le t \le 4 \\ a &= 19-5e^{8-2t} \quad \text{for } 4 \le t \le 10 \end{aligned} $$
When $t=3$, the velocity of $P$ is $-\frac{1}{3}\text{ ms}^{-1}$ and its displacement from $O$ is $-\frac{1}{4}\text{ m}$.
-
- Find the velocity of $P$ when $t=4$.
- Find the displacement of $P$ from $O$ when $t=4$.
- Find the displacement of $P$ from $O$ when $t=10$.
(a)(i)
For $0 \le t \le 4$:
$$ \begin{aligned} v &= \int (t^2 - 2) \text{ d}t = \frac{t^3}{3} - 2t + c \end{aligned} $$
When $t=3, v = -1/3$:
$$ \begin{aligned} -\frac{1}{3} &= \frac{27}{3} - 6 + c \implies -\frac{1}{3} = 9 - 6 + c \implies c = -\frac{10}{3} \\ \therefore\quad v &= \frac{t^3}{3} - 2t - \frac{10}{3} \end{aligned} $$
When $t=4$:
$$ \begin{aligned} v(4) &= \frac{64}{3} - 8 - \frac{10}{3} = \frac{54}{3} - 8 = 18 - 8 = 10 \text{ ms}^{-1} \end{aligned} $$
(a)(ii)
$$ \begin{aligned} s &= \int \left( \frac{t^3}{3} - 2t - \frac{10}{3} \right) \text{ d}t = \frac{t^4}{12} - t^2 - \frac{10t}{3} + d \end{aligned} $$
When $t=3, s = -1/4$:
$$ \begin{aligned} -\frac{1}{4} &= \frac{81}{12} - 9 - 10 + d \\ -\frac{1}{4} &= \frac{27}{4} - 19 + d \implies d = 19 - \frac{28}{4} = 19 - 7 = 12 \\ \therefore\quad s &= \frac{t^4}{12} - t^2 - \frac{10t}{3} + 12 \end{aligned} $$
When $t=4$:
$$ \begin{aligned} s(4) &= \frac{256}{12} - 16 - \frac{40}{3} + 12 = \frac{64}{3} - 4 - \frac{40}{3} = \frac{24}{3} - 4 = 8 - 4 = 4 \text{ m} \end{aligned} $$
(b)
For $4 \le t \le 10$, $a = 19 - 5e^{8-2t}$:
$$ \begin{aligned} v &= \int (19 - 5e^{8-2t}) \text{ d}t = 19t + \frac{5}{2}e^{8-2t} + E \end{aligned} $$
At $t=4, v=10$:
$$ \begin{aligned} 10 &= 19(4) + 2.5e^0 + E \implies 10 = 76 + 2.5 + E \implies E = -68.5 \end{aligned} $$
$$ \begin{aligned} s &= \int (19t + 2.5e^{8-2t} - 68.5) \text{ d}t = 9.5t^2 - 1.25e^{8-2t} - 68.5t + F \end{aligned} $$
At $t=4, s=4$:
$$ \begin{aligned} 4 &= 9.5(16) - 1.25 - 68.5(4) + F \\ 4 &= 152 - 1.25 - 274 + F \implies 4 = -123.25 + F \implies F = 127.25 \end{aligned} $$
When $t=10$:
$$ \begin{aligned} s(10) &= 9.5(100) - 1.25e^{-12} - 68.5(10) + 127.25 \\ &= 950 - 685 + 127.25 - 1.25e^{-12} \\ &= 392.25 - 1.25e^{-12} \text{ m} \end{aligned} $$
49.
[Source: 0606/22/M/J/25 - Question No. 2]
In this question, all lengths are in metres and time is in seconds. A particle $P$ moves in a straight line such that its displacement $s$ from a fixed point $O$ at time $t$ is given by $s=(t-4)^{2}(t-1)$ for $t\ge0$.
- On the axes, sketch the displacement-time graph of $P$, stating the intercepts with the axes.
- Find an expression for the velocity, $v$, of $P$. Give your answer in a factorised form.
- On the axes, sketch the velocity-time graph of $P$, stating the intercepts with the axes.
- Find an expression for the acceleration, $a$, of $P$.
- On the axes, sketch the acceleration-time graph of $P$, stating the intercepts with the axes.
(a)
Intercepts: $t$-axis at $t=1$ and $t=4$ (touching). $s$-axis at $s=-16$.
(b)
$$ \begin{aligned} s &= (t-4)^2(t-1) \\ v &= \frac{\text{d}s}{\text{d}t} = 2(t-4)(t-1) + (t-4)^2 \\ &= (t-4) [ 2(t-1) + (t-4) ] \\ &= (t-4) (2t - 2 + t - 4) \\ &= (t-4)(3t - 6) = 3(t-2)(t-4) \end{aligned} $$
(c)
Intercepts: $t$-axis at $t=2$ and $t=4$. $v$-axis at $v=24$. Minimum at $t=3, v=-3$.
(d)
$$ \begin{aligned} v &= 3(t^2 - 6t + 8) = 3t^2 - 18t + 24 \\ a &= \frac{\text{d}v}{\text{d}t} = 6t - 18 = 6(t-3) \end{aligned} $$
(e)
Intercepts: $t$-axis at $t=3$. $a$-axis at $a=-18$.
50.
[Source: 0606/11/O/N/25 - Question No. 12]
The diagram shows the velocity-time graph for a particle moving in a straight line. It is given that the velocity-time graph is part of a quadratic curve. The curve has gradient $0$ when time is $0$.
- Sketch the speed-time graph for the motion of this particle.
- Sketch a possible acceleration-time graph for the motion of the particle.
(a)
Given the curve is quadratic with gradient $0$ at $t=0$, the velocity function is of the form $v = a t^2 + c$. The graph shows $c < 0$ and $a > 0$.
Speed is the absolute value of velocity, $|v|$. The negative portion of the graph is reflected across the $t$-axis.
Speed is the absolute value of velocity, $|v|$. The negative portion of the graph is reflected across the $t$-axis.
(b)
Acceleration $a = \frac{\text{d}v}{\text{d}t} = 2at$. This is a straight line passing through the origin with a positive gradient.
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